MathsCBSE Class 9 · Chapter 4

Exploring Algebraic Identities Class 9: Identities, Factorisation and Solved Questions

Q: Which identities must I learn for Class 9 (2026-27)?

The NCERT chapter summary lists eleven: (x \pm y)^2, (x + y + z)^2, (x + y)(x - y), (x + a)(x + b), (ax + b)(cx + d), x^3 \pm y^3, (x \pm y)^3 and x^3 + y^3 + z^3 - 3xyz. Learn all of them with one example each.

Q: Do I need to draw algebra tiles in the exam?

The CBSE syllabus includes visualising factorisation with algebra tiles, so you may be asked to show or explain a tile arrangement, such as the rectangle for x^2 + 7x + 12. Practise drawing one or two neatly.

Q: Is factorising by the factor theorem part of this chapter?

No. In the 2026-27 syllabus, quadratics are factorised by identities and splitting the middle term. The factor theorem is not listed.

Q: How do I know which identity to use?

Count the terms and look at the powers: two square terms with a minus between them suggest a^2 - b^2; three terms with squares at both ends suggest a perfect square; two cube terms suggest a sum or difference of cubes; four terms with cubes at both ends suggest (a \pm b)^3.

Q: Why do I have to say the denominator is not zero when simplifying?

Cancelling a factor means dividing by it, which is only allowed when it is not zero. The simplified expression is equal to the original only where the original is defined.

Updated 15 October 2026 · Fact-checked

Exploring Algebraic Identities in Class 9 teaches identities, equations true for every value of the variables, such as (a + b)² = a² + 2ab + b². You see them as areas and volumes, use them for quick squares and cubes, factorise expressions including quadratics by splitting the middle term, and simplify rational expressions.

What Exploring Algebraic Identities is about

This chapter of *Ganita Manjari* Part I matches CBSE's Exploring Algebraic Identities (Unit II, Algebra). It opens with a puzzle: take three consecutive square numbers, add the smallest and largest, subtract twice the middle one, and you always get 2. By the end you can prove why.

The chapter builds a toolkit of identities and shows why each one is true with pictures: a square of side $a + b$ cut into $a^{2}$ , $b^{2}$ and two $ab$ rectangles; a cube of edge $a + b$ cut into two cubes and six cuboids; algebra tiles arranged into rectangles.

What you must be able to do:

expand squares and cubes of binomials and the square of a trinomial
use identities for mental arithmetic, such as $9 8^{2}$ or $103 \times 97$
factorise perfect squares, differences of squares, sums and differences of cubes, trinomial squares and quadratics like $6 x^{2} - 7 x - 20$ (by splitting the middle term)
use $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
simplify rational expressions by cancelling common factors

This is a high-scoring chapter if your expansions are careful with signs. Questions run from 1-mark MCQs to 5-mark problems where an area or volume has to be factorised into lengths.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 marks in the school annual exam, shared with Introduction to Polynomials, Sequences and Progressions and Linear Equations in Two Variables.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit II Algebra

Key concepts

Identity or equation?

NCERT section 4.2 Visualising Identities

An algebraic identity is an equality that is true for every value of its variables. An equation is true only for some values.

$x^{2} - 1 = 24$ is an equation: only $x = 5$ or $x = - 5$ work.
$(x + y)^{2} = x^{2} + 2 x y + y^{2}$ is an identity: it works for all x and y, including negatives and fractions.

A single example cannot prove an identity, but a single counter-example is enough to show something is not an identity. To prove an identity, expand using the distributive property.

Note that $(a + b)^{2} \neq = a^{2} + b^{2}$ in general; the difference is $2 ab$ , so $(a + b)^{2}$ is bigger when $ab > 0$ , smaller when $ab < 0$ , and equal only when $ab = 0$ .

Worked example

(i) Is $(x + 1)^{2} = x^{2} + 1$ an identity? (ii) For $a = 7$ , $b = - 3$ , compare $(a + b)^{2}$ with $a^{2} + b^{2}$ and explain the difference.

Try it first, then show the solutionSolution

(i) Try $x = 1$ : left side $= 4$ , right side $= 2$ . They differ, so it is not an identity.
It is an equation: $x^{2} + 2 x + 1 = x^{2} + 1$ gives $2 x = 0$ , so it holds only for $x = 0$ .
(ii) $(a + b)^{2} = 4^{2} = 16$ and $a^{2} + b^{2} = 49 + 9 = 58$ .
The difference is $2 ab = 2 (7) (- 3) = - 42$ , and $58 - 42 = 16$ . Since $ab < 0$ , $(a + b)^{2}$ is smaller.

Answer: (i) Not an identity; true only for $x = 0$ . (ii) $16 < 58$ because $2 ab = - 42$ .

Square of a sum and of a difference

NCERT section 4.2 Visualising Identities

$(a + b)^{2} = a^{2} + 2 ab + b^{2} (a - b)^{2} = a^{2} - 2 ab + b^{2}$

Picture: a square of side $a + b$ splits into a square $a^{2}$ , a square $b^{2}$ and two rectangles $ab$ , which gives the first identity. The second comes from replacing b by $- b$ , or from cutting the two rectangles off a square of side a.

Uses:

Expanding: $(5 x + 2 y)^{2} = 25 x^{2} + 20 x y + 4 y^{2}$ . Square each term including its coefficient.
Mental arithmetic: write the number as a round number plus or minus a small one, e.g. $4 3^{2} = (40 + 3)^{2} = 1849$ .
Proving patterns: $(n - 1)^{2} + (n + 1)^{2} - 2 n^{2} = 2$ , which explains the opening puzzle.

Worked example

(i) Expand $(3 p - 5 q)^{2}$ . (ii) Find $9 8^{2}$ and $5 2^{2}$ using identities.

Try it first, then show the solutionSolution

(i) $(3 p)^{2} - 2 (3 p) (5 q) + (5 q)^{2} = 9 p^{2} - 30 pq + 25 q^{2}$ .
(ii) $9 8^{2} = (100 - 2)^{2} = 10000 - 400 + 4 = 9604$ .
$5 2^{2} = (50 + 2)^{2} = 2500 + 200 + 4 = 2704$ .

Answer: (i) $9 p^{2} - 30 pq + 25 q^{2}$ (ii) $9604$ and $2704$

Factorising perfect squares

NCERT section 4.3 Factorisation of Algebraic Expressions Using Identities

Reading an identity from right to left factorises. An expression is a perfect square if:
1. the first and last terms are squares, $A^{2}$ and $B^{2}$ (after taking out any common factor), and
2. the middle term is $\pm 2 A B$ .

Then $A^{2} + 2 A B + B^{2} = (A + B)^{2}$ and $A^{2} - 2 A B + B^{2} = (A - B)^{2}$ .

Always look for a common factor first: $50 p^{2} + 60 pq + 18 q^{2} = 2 (25 p^{2} + 30 pq + 9 q^{2}) = 2 (5 p + 3 q)^{2}$ .

Worked example

Factorise (i) $49 x^{2} - 42 x y + 9 y^{2}$ (ii) $12 a^{2} + 60 ab + 75 b^{2}$ .

Try it first, then show the solutionSolution

(i) $49 x^{2} = (7 x)^{2}$ , $9 y^{2} = (3 y)^{2}$ , and $2 (7 x) (3 y) = 42 x y$ with a minus sign.
So $49 x^{2} - 42 x y + 9 y^{2} = (7 x - 3 y)^{2}$ .
(ii) Common factor 3: $3 (4 a^{2} + 20 ab + 25 b^{2})$ .
$4 a^{2} = (2 a)^{2}$ , $25 b^{2} = (5 b)^{2}$ , $2 (2 a) (5 b) = 20 ab$ , so it is $3 (2 a + 5 b)^{2}$ .

Answer: (i) $(7 x - 3 y)^{2}$ (ii) $3 (2 a + 5 b)^{2}$

Square of a trinomial

NCERT section 4.4 More Identities

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$

It follows from $(a + d)^{2}$ with $d = b + c$ . In the picture, a square of side $a + b + c$ splits into three squares and six rectangles.

When terms are negative, put the sign inside the letter: for $(2 x - y + 3 z)^{2}$ take $a = 2 x$ , $b = - y$ , $c = 3 z$ . Each product term picks up the signs of the two terms it is made from.

To factorise six-term expressions like $4 a^{2} + 9 b^{2} + c^{2} - 12 ab + 6 b c - 4 a c$ , find the square roots of the three square terms and choose signs so that every product term matches.

Worked example

(i) Expand $(2 x - y + 3 z)^{2}$ . (ii) Find $21 3^{2}$ using this identity.

Try it first, then show the solutionSolution

(i) Squares: $4 x^{2} + y^{2} + 9 z^{2}$ .
Products: $2 (2 x) (- y) = - 4 x y$ , $2 (- y) (3 z) = - 6 y z$ , $2 (3 z) (2 x) = 12 z x$ .
$(2 x - y + 3 z)^{2} = 4 x^{2} + y^{2} + 9 z^{2} - 4 x y - 6 y z + 12 z x$ .
(ii) $213 = 200 + 10 + 3$ : $40000 + 100 + 9 + 2 (2000 + 30 + 600) = 40109 + 5260 = 45369$ .

Answer: (i) $4 x^{2} + y^{2} + 9 z^{2} - 4 x y - 6 y z + 12 z x$ (ii) $45369$

Difference of squares and Śrīdharāchārya's method

NCERT section 4.4 More Identities

$a^{2} - b^{2} = (a + b) (a - b)$

This gives fast products of numbers equally spaced around a round number: $103 \times 97 = 10 0^{2} - 3^{2} = 9991$ .

Rewritten as $a^{2} = (a + b) (a - b) + b^{2}$ , it was used by Śrīdharāchārya (c. 750 CE) to square numbers quickly. Choose b to make $a + b$ and $a - b$ easy: $7 5^{2} = 80 \times 70 + 5^{2} = 5625$ . Numbers ending in 5 always work nicely this way.

As a factorising tool: $36 s^{2} - 49 t^{2} = (6 s + 7 t) (6 s - 7 t)$ .

Worked example

Use identities to find (i) $103 \times 97$ (ii) $7 5^{2}$ by Śrīdharāchārya's method (iii) factorise $36 s^{2} - 49 t^{2}$ .

Try it first, then show the solutionSolution

(i) $(100 + 3) (100 - 3) = 10000 - 9 = 9991$ .
(ii) Take $a = 75$ , $b = 5$ : $7 5^{2} = 80 \times 70 + 25 = 5600 + 25 = 5625$ .
(iii) $(6 s)^{2} - (7 t)^{2} = (6 s + 7 t) (6 s - 7 t)$ .

Answer: (i) 9991 (ii) 5625 (iii) $(6 s + 7 t) (6 s - 7 t)$

Factorising quadratics: algebra tiles and splitting the middle term

NCERT section 4.6 Factorisation Without Using Algebra Tiles

$(x + a) (x + b) = x^{2} + (a + b) x + ab$ . So to factorise $x^{2} + p x + q$ , find two numbers whose sum is p and product is q.

With algebra tiles, $x^{2} + 7 x + 12$ is one $x^{2}$ -tile, seven x-tiles and twelve unit tiles. They make a rectangle only when 7x is split as $3 x + 4 x$ and the units form a $3 \times 4$ block, showing sides $x + 3$ and $x + 4$ .

For $a x^{2} + b x + c$ with $a \neq = 1$ , find two numbers with sum b and product ac, split the middle term with them, and group in pairs. Sign guide:

c positive, b positive: both numbers positive
c positive, b negative: both negative
c negative: one positive and one negative

Worked example

Factorise (i) $x^{2} + 2 x - 35$ (ii) $6 x^{2} + 11 x + 3$ .

Try it first, then show the solutionSolution

(i) Need sum 2 and product $- 35$ : the numbers are 7 and $- 5$ .
$x^{2} + 2 x - 35 = (x + 7) (x - 5)$ .
(ii) $a c = 18$ , $b = 11$ : the numbers are 9 and 2.
$6 x^{2} + 9 x + 2 x + 3 = 3 x (2 x + 3) + 1 (2 x + 3) = (3 x + 1) (2 x + 3)$ .
Check (ii) at $x = 5$ : $6 (25) + 55 + 3 = 208$ and $16 \times 13 = 208$ .

Answer: (i) $(x + 7) (x - 5)$ (ii) $(3 x + 1) (2 x + 3)$

Cube of a sum and of a difference

NCERT section 4.7 Finding New Identities

$(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} (a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$

Picture: a cube of edge $a + b$ splits into a cube $a^{3}$ , a cube $b^{3}$ , three cuboids $a \times a \times b$ and three cuboids $a \times b \times b$ .

In $(a - b)^{3}$ the signs alternate $+, -, +, -$ . A short form that is handy for values: $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$ .

Use them for cubes of numbers near round numbers ( $10 2^{3}$ , $9 9^{3}$ ) and to recognise a four-term expression as a cube, which gives the edge of a cube from its volume.

Worked example

(i) Find $10 2^{3}$ and $9 9^{3}$ . (ii) Factorise $27 p^{3} - 54 p^{2} q + 36 p q^{2} - 8 q^{3}$ .

Try it first, then show the solutionSolution

(i) $10 2^{3} = (100 + 2)^{3} = 1000000 + 3 (10000) (2) + 3 (100) (4) + 8 = 1061208$ .
$9 9^{3} = (100 - 1)^{3} = 1000000 - 30000 + 300 - 1 = 970299$ .
(ii) $27 p^{3} = (3 p)^{3}$ and $8 q^{3} = (2 q)^{3}$ ; check $3 (3 p)^{2} (2 q) = 54 p^{2} q$ and $3 (3 p) (2 q)^{2} = 36 p q^{2}$ .
Signs alternate, so it is $(3 p - 2 q)^{3}$ .

Answer: (i) 1061208 and 970299 (ii) $(3 p - 2 q)^{3}$

Sum and difference of cubes, and x³ + y³ + z³ − 3xyz

NCERT section 4.7 Finding New Identities

$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2}) x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$
$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

All three are checked by multiplying out. Two useful results follow from the last one:

If $x + y + z = 0$ , then $x^{3} + y^{3} + z^{3} = 3 x y z$ .
If you know $x + y + z$ , $x y + y z + z x$ and $x y z$ , you can find $x^{3} + y^{3} + z^{3}$ without knowing x, y, z. Use $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$ to get the sum of squares first.

Worked example

(i) Factorise $8 m^{3} - 27$ . (ii) Without cubing, find $(- 12)^{3} + 7^{3} + 5^{3}$ .

Try it first, then show the solutionSolution

(i) $8 m^{3} - 27 = (2 m)^{3} - 3^{3} = (2 m - 3) (4 m^{2} + 6 m + 9)$ .
(ii) $- 12 + 7 + 5 = 0$ , so the sum of cubes is $3 x y z$ .
$3 (- 12) (7) (5) = - 1260$ .
Check: $- 1728 + 343 + 125 = - 1260$ .

Answer: (i) $(2 m - 3) (4 m^{2} + 6 m + 9)$ (ii) $- 1260$

Simplifying rational expressions

NCERT section 4.8 Simplifying Rational Expressions

A rational expression is one polynomial divided by another. To simplify:
1. Factorise the numerator and the denominator completely.
2. Cancel factors that are common to both, assuming the cancelled factor is not zero.
3. State the values of the variable for which the original denominator is zero (the expression is not defined there).

You may cancel only factors (things multiplied), never terms (things added): $\frac{x + 3}{x + 5}$ cannot be simplified.

Worked example

Simplify $\frac{x ^{2} - 9}{x ^{2} + x - 6}$ , assuming the denominator is not zero.

Try it first, then show the solutionSolution

Numerator: $x^{2} - 9 = (x + 3) (x - 3)$ .
Denominator: sum 1, product $- 6$ gives 3 and $- 2$ : $(x + 3) (x - 2)$ .
Cancel the common factor $(x + 3)$ (it is not zero, since the denominator is not zero).
Result: $\frac{x - 3}{x - 2}$ , valid for $x \neq = - 3$ and $x \neq = 2$ .

Answer: $\frac{x - 3}{x - 2}$

Maths in real life: Exploring Algebraic Identities

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Square of a sum

Suppose a square vegetable patch of side 20 m is made larger by adding 3 m to each side length.

By how much does the area increase?

Try it first, then show the solutionSolution

New area $= (20 + 3)^{2} = 400 + 120 + 9 = 529$ m².
Increase $= 529 - 400 = 129$ m², which is the two $20 \times 3$ strips plus the $3 \times 3$ corner.

Answer: 129 m²

Memory hook: The extra area is two strips and a corner: $2 ab + b^{2}$ .

Difference of squares

Suppose a shopkeeper sells 48 pens at ₹52 each and wants the total quickly.

Find the total without long multiplication.

Try it first, then show the solutionSolution

$48 \times 52 = (50 - 2) (50 + 2)$ .
$= 2500 - 4 = 2496$ .

Answer: ₹2,496

Memory hook: Numbers equally spaced around a round number: square the middle, subtract the gap squared.

Factorising a quadratic

Suppose a hall's floor area is $x^{2} + 7 x + 10$ square metres, where x is a length fixed by the architect.

Find the length and breadth in terms of x, and their values when $x = 8$ .

Try it first, then show the solutionSolution

Sum 7, product 10: 2 and 5, so $(x + 2) (x + 5)$ .
At $x = 8$ : 10 m by 13 m.

Answer: $(x + 5)$ m by $(x + 2)$ m; 13 m by 10 m when $x = 8$ .

Memory hook: An area expression hides its sides inside its factors.

Area of a border

Suppose a square park of side 40 m has a 2 m wide walking path running along the inside of its boundary.

Find the area of the path.

Try it first, then show the solutionSolution

Inner square side $= 40 - 4 = 36$ m.
Path area $= 4 0^{2} - 3 6^{2} = (40 + 36) (40 - 36) = 76 \times 4 = 304$ m².

Answer: 304 m²

Memory hook: Border area = big square minus small square = (sum)(difference).

Cube of a sum

Suppose a cube-shaped storage box of edge 10 cm is redesigned with edge 11 cm.

How much extra volume does the new box hold?

Try it first, then show the solutionSolution

$1 1^{3} = (10 + 1)^{3} = 1000 + 300 + 30 + 1 = 1331$ cm³.
Extra $= 1331 - 1000 = 331$ cm³.

Answer: 331 cm³

Memory hook: Grow a cube by 1 and you add three slabs, three rods and one tiny cube.

Topics in this chapter

Exploring Algebraic Identities formula sheet

Square of a sum: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$
Square of a difference: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$
Square of a trinomial: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$
Difference of squares: $a^{2} - b^{2} = (a + b) (a - b)$; Śrīdharāchārya's form: $a^{2} = (a + b) (a - b) + b^{2}$ .
Product of two binomials: $(x + a) (x + b) = x^{2} + (a + b) x + ab$
General product: $(a x + b) (c x + d) = a c x^{2} + (a d + b c) x + b d$; Splitting the middle term: find numbers with sum $a d + b c$ and product $a c b d$ .
Cube of a sum: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = a^{3} + b^{3} + 3 ab (a + b)$
Cube of a difference: $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} = a^{3} - b^{3} - 3 ab (a - b)$
Difference of cubes: $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$
Sum of cubes: $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$
Three cubes: $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$
Special case: $x + y + z = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z$
Sum of squares from sums: $x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2 (x y + y z + z x)$
Reciprocal pairs: $x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2, x^{3} + \frac{1}{x ^{3}} = (x + \frac{1}{x})^{3} - 3 (x + \frac{1}{x})$; Special cases of $(a + b)^{2}$ and $(a + b)^{3}$ with $ab = 1$ .

How the school exam asks Exploring Algebraic Identities

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Expand using an identity	Expand $(- 3 x + 4)^{2}$ , $(2 s + 7) (2 s - 7)$ , $(p + 3 q + 7 r)^{2}$ or $(2 a - 3 b)^{3}$ .	Name a, b (and c) with their signs, write the identity, substitute in brackets, then simplify each term. Count terms: 3 for a binomial square, 6 for a trinomial square, 4 for a cube.
Mental arithmetic with identities	Find $99 8^{2}$ , $104 \times 96$ , $19 9^{3}$ or $7 5^{2}$ without direct multiplication.	Rewrite around a round number. Products of numbers equally spaced from a round number use $a^{2} - b^{2}$ ; squares use $(a \pm b)^{2}$ ; cubes use $(a \pm b)^{3}$ .
Factorise using identities	Perfect squares, differences of squares, sums and differences of cubes, trinomial squares, and four-term cubes.	Take out common factors first. Identify the square or cube roots of the end terms, then check the middle terms against the identity, including signs.
Factorise a quadratic	Factorise $x^{2} - 13 x + 42$ , $6 x^{2} - 7 x - 20$ , or find the sides of a rectangle with area $2 x^{2} + 7 x + 3$ .	Find two numbers with the right sum and product (product ac when the leading coefficient is not 1). Split, group, take out the common bracket, and check by expanding.
Values from sums and products	Given $a + b + c$ and $ab + b c + c a$ (and perhaps abc), find $a^{2} + b^{2} + c^{2}$ or $a^{3} + b^{3} + c^{3}$ ; or given $x + \frac{1}{x}$ , find $x^{2} + \frac{1}{x ^{2}}$ .	Write the identity that links what is given to what is asked, substitute, and solve. Do not try to find the individual numbers.
Simplify a rational expression	Simplify $\frac{x ^{2} - 7 x + 12}{5 x ^{2} + 5 x - 100}$ , assuming the denominator is not zero.	Factorise top and bottom fully, cancel only common factors, and state the excluded values.

Tricks to remember Exploring Algebraic Identities

Short memory aids. Say each one aloud once, then use it on a question the same day.

First, twice the product, last
Square of a binomial: square the first, twice the product, square the last. Only the middle sign follows the bracket.
Use it when: Expanding $(a \pm b)^{2}$ without dropping the $2 ab$ .
1-3-3-1
The coefficients of a binomial cube are 1, 3, 3, 1, with powers of a going down 3, 2, 1, 0 and powers of b going up 0, 1, 2, 3.
Use it when: Writing $(a + b)^{3}$ and $(a - b)^{3}$ ; for minus, the signs alternate.
SOAP for cubes
Sum or difference of cubes: the first bracket has the Same sign, the second bracket starts with the Opposite sign, and the last term is Always Positive.
Use it when: Factorising $x^{3} \pm y^{3}$ without mixing up signs.
Sum and product hunt
For $x^{2} + p x + q$ , hunt for a pair: they ADD to p and MULTIPLY to q. The sign of q tells you same signs (+) or opposite signs (−).
Use it when: Fast factorising of quadratics.
Cancel only what is multiplied
Factors can leave the fraction together; terms joined by + or − must stay.
Use it when: Simplifying rational expressions correctly.

Exploring Algebraic Identities: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Difference of two squares of binomials · 1 mark · MCQExam level

$(a - b)^{2} - (a + b)^{2}$ equals
(a) 0
(b) $- 2 ab$
(c) $- 4 ab$
(d) $2 b^{2}$

Try it first, then show the solutionSolution

$(a^{2} - 2 ab + b^{2}) - (a^{2} + 2 ab + b^{2}) = - 4 ab$ .

Answer: (c) $- 4 ab$

Example 2: Which is an identity? · 1 mark · MCQExam level

Which of the following is an identity?
(a) $x^{2} - 1 = 24$
(b) $(x + 2)^{2} = x^{2} + 4$
(c) $(x - 3) (x + 3) = x^{2} - 9$
(d) $2 x + 1 = 7$

Try it first, then show the solutionSolution

(a) and (d) hold for particular x only. (b) fails at $x = 1$ ( $9 \neq = 5$ ).
(c) is $a^{2} - b^{2} = (a + b) (a - b)$ , true for all x.

Answer: (c)

Example 3: Squaring 999 (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): $99 9^{2} = 998001$ .
Reason (R): $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$99 9^{2} = (1000 - 1)^{2} = 1000000 - 2000 + 1 = 998001$ , using R.

Answer: (a)

Example 4: Perfect square · 2 marks · Very short answerExam level

Factorise $25 x^{2} + 40 x y + 16 y^{2}$ .

Try it first, then show the solutionSolution

$25 x^{2} = (5 x)^{2}$ , $16 y^{2} = (4 y)^{2}$ , $2 (5 x) (4 y) = 40 x y$ .
So it is $(5 x + 4 y)^{2}$ .

Answer: $(5 x + 4 y)^{2}$

Example 5: Trinomial square in reverse · 3 marks · Short answerHarder exam level

Factorise $4 a^{2} + 9 b^{2} + c^{2} - 12 ab + 6 b c - 4 a c$ .

Try it first, then show the solutionSolution

Roots of the square terms: $2 a$ , $3 b$ , $c$ .
$- 12 ab$ and $- 4 a c$ mean a has the opposite sign to both b and c; $+ 6 b c$ means b and c have the same sign.
Take $2 a$ , $- 3 b$ , $- c$ .
Check: $2 (2 a) (- 3 b) = - 12 ab$ , $2 (- 3 b) (- c) = 6 b c$ , $2 (- c) (2 a) = - 4 a c$ .

Answer: $(2 a - 3 b - c)^{2}$

Example 6: Products without multiplying · 2 marks · Very short answerHarder exam level

Using identities, evaluate (i) $107 \times 93$ (ii) $99 8^{2}$ .

Try it first, then show the solutionSolution

(i) $(100 + 7) (100 - 7) = 10000 - 49 = 9951$ .
(ii) $(1000 - 2)^{2} = 1000000 - 4000 + 4 = 996004$ .

Answer: (i) 9951 (ii) 996004

Example 7: Splitting the middle term · 3 marks · Short answerHarder exam level

Factorise $6 x^{2} - 7 x - 20$ .

Try it first, then show the solutionSolution

$a c = 6 \times (- 20) = - 120$ ; need two numbers with product $- 120$ and sum $- 7$ : $- 15$ and 8.
$6 x^{2} - 15 x + 8 x - 20$ .
$= 3 x (2 x - 5) + 4 (2 x - 5) = (2 x - 5) (3 x + 4)$ .
Check at $x = 3$ : $54 - 21 - 20 = 13$ and $(1) (13) = 13$ .

Answer: $(2 x - 5) (3 x + 4)$

Example 8: Reciprocal pairs · 3 marks · Short answerHarder exam level

If $x + \frac{1}{x} = 5$ , find $x^{2} + \frac{1}{x ^{2}}$ and $x^{3} + \frac{1}{x ^{3}}$ .

Try it first, then show the solutionSolution

$(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x ^{2}}$ , so $x^{2} + \frac{1}{x ^{2}} = 25 - 2 = 23$ .
$(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x ^{3}} + 3 (x + \frac{1}{x})$ .
$125 = x^{3} + \frac{1}{x ^{3}} + 15$ , so $x^{3} + \frac{1}{x ^{3}} = 110$ .

Answer: 23 and 110

Example 9: Simplifying a rational expression · 3 marks · Short answerHarder exam level

Simplify $\frac{2 x ^{2} + 5 x - 3}{4 x ^{2} - 1}$ and state the values of x for which the original expression is not defined.

Try it first, then show the solutionSolution

Numerator: $(2 x - 1) (x + 3)$ (split $5 x$ as $6 x - x$ ).
Denominator: $(2 x - 1) (2 x + 1)$ .
Cancel $(2 x - 1)$ : $\frac{x + 3}{2 x + 1}$ .
Not defined when $4 x^{2} - 1 = 0$ : $x = \frac{1}{2}$ or $x = - \frac{1}{2}$ .

Answer: $\frac{x + 3}{2 x + 1}$ ; not defined at $x = \pm \frac{1}{2}$ .

Example 10: Edge of a cube from its volume · 5 marks · Long answerHarder exam level

The volume of a cube is $8 x^{3} + 36 x^{2} y + 54 x y^{2} + 27 y^{3}$ cubic units. (i) Find its edge. (ii) Find the edge and volume when $x = 1$ and $y = 2$ , and check by substituting in the given expression. (iii) Find its total surface area in terms of x and y.

Try it first, then show the solutionSolution

$8 x^{3} = (2 x)^{3}$ , $27 y^{3} = (3 y)^{3}$ ; $3 (2 x)^{2} (3 y) = 36 x^{2} y$ and $3 (2 x) (3 y)^{2} = 54 x y^{2}$ .
So the volume is $(2 x + 3 y)^{3}$ and the edge is $2 x + 3 y$ units.
For $x = 1$ , $y = 2$ : edge $= 8$ , volume $= 512$ .
Check: $8 + 72 + 216 + 216 = 512$ .
Total surface area $= 6 (2 x + 3 y)^{2} = 6 (4 x^{2} + 12 x y + 9 y^{2}) = 24 x^{2} + 72 x y + 54 y^{2}$ square units.

Answer: (i) $2 x + 3 y$ (ii) edge 8, volume 512 (iii) $24 x^{2} + 72 x y + 54 y^{2}$

Example 11: Sum of cubes from sums · 3 marks · Short answerHOTS / competency

If $a + b + c = 9$ and $ab + b c + c a = 26$ , find $a^{2} + b^{2} + c^{2}$ and $a^{3} + b^{3} + c^{3} - 3 ab c$ .

Try it first, then show the solutionSolution

$a^{2} + b^{2} + c^{2} = 81 - 2 (26) = 29$ .
$a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - (ab + b c + c a))$ .
$= 9 (29 - 26) = 27$ .

Answer: 29 and 27

Example 12: Path around a playground · 5 marks · Long answerHOTS / competency

A square playground has side 30 m. A path of uniform width s m is laid all around the outside. (i) Show that the area of the path is $4 s^{2} + 120 s$ square metres. (ii) If the path covers 256 m², find s.

Try it first, then show the solutionSolution

(i) Outer square side $= 30 + 2 s$ , so path area $= (30 + 2 s)^{2} - 3 0^{2}$ .
$= 900 + 120 s + 4 s^{2} - 900 = 4 s^{2} + 120 s$ .
(ii) $4 s^{2} + 120 s = 256$ , so $s^{2} + 30 s - 64 = 0$ .
Numbers with product $- 64$ and sum 30: 32 and $- 2$ , so $(s + 32) (s - 2) = 0$ .
Width cannot be negative, so $s = 2$ m. Check: $3 4^{2} - 900 = 256$ .

Answer: (i) shown (ii) $s = 2$ m

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Saira's algebra tiles

Saira has square tiles of side x units ( $x^{2}$ -tiles), rectangular strips of x units by 1 unit (x-tiles) and unit squares. She arranges some of them, without gaps or overlaps, into one large rectangle each time.

(i) · 1 mark
Which tiles does she need to make a rectangle of sides $x + 2$ and $x + 6$ ?
Answer
Answer: One $x^{2}$ -tile, 8 x-tiles and 12 unit tiles
$(x + 2) (x + 6) = x^{2} + 8 x + 12$ .
(ii) · 1 mark
She uses one $x^{2}$ -tile, 9 x-tiles and 20 unit tiles. Find the sides of her rectangle.
Answer
Answer: $x + 4$ and $x + 5$
$x^{2} + 9 x + 20$ : sum 9, product 20 gives 4 and 5, so $(x + 4) (x + 5)$ .
(iii) · 2 marks
If $x = 7$ cm, find the area and the perimeter of the rectangle in part (ii), and check the area from the tile count.
Answer
Answer: Area 132 cm², perimeter 46 cm
Sides 11 cm and 12 cm, so area $= 132$ cm² and perimeter $= 2 (11 + 12) = 46$ cm (1 mark). Tile check: $49 + 9 \times 7 + 20 = 132$ (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Expand $(3 x - 4 y)^{2}$ . (2 marks)

Step	Marks
$(3 x)^{2} - 2 (3 x) (4 y) + (4 y)^{2}$ , correct identity with correct a and b	1
$= 9 x^{2} - 24 x y + 16 y^{2}$	1

The area of a rectangle is $2 x^{2} + 7 x + 3$ square units. Find expressions for its length and breadth, and the perimeter. (3 marks)

Step	Marks
$a c = 6$ , $b = 7$ : numbers 6 and 1; $2 x^{2} + 6 x + x + 3$	1
$2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3)$ , so the sides are $2 x + 1$ and $x + 3$	1
Perimeter $= 2 (2 x + 1 + x + 3) = 6 x + 8$ units	1

If $a + b + c = 6$ , $ab + b c + c a = 11$ and $ab c = 6$ , find $a^{2} + b^{2} + c^{2}$ and $a^{3} + b^{3} + c^{3}$ . (5 marks)

Step	Marks
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$	1
$36 = a^{2} + b^{2} + c^{2} + 22$ , so $a^{2} + b^{2} + c^{2} = 14$	1
$a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$	1
$= 6 (14 - 11) = 18$	1
$a^{3} + b^{3} + c^{3} = 18 + 3 \times 6 = 36$	1

Where students lose marks in Exploring Algebraic Identities

The slips examiners see most often, and what to write instead.

Mistake: Writing (a + b)² as a² + b²
The middle term $2 ab$ is forgotten.
Wrong working
$(x + 5)^{2} = x^{2} + 25$
Right working
$(x + 5)^{2} = x^{2} + 10 x + 25$
Do this: Always write three terms for a binomial square; test with a number if unsure.
Usually costs: The full mark for the expansion.
Mistake: Not squaring the coefficient
Only the letter is squared.
Wrong working
$(3 x - 2)^{2} = 3 x^{2} - 12 x + 4$
Right working
$(3 x - 2)^{2} = 9 x^{2} - 12 x + 4$
Do this: Put the whole term in a bracket before squaring: $(3 x)^{2}$ .
Usually costs: 1 mark.
Mistake: Wrong signs in a trinomial square
Negative terms are not tracked in the product terms.
Wrong working
$(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 ab - 2 b c - 2 c a$
Right working
$(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 ab + 2 b c - 2 c a$ (since $(- b) (- c) = + b c$ )
Do this: Write $a$ , $- b$ , $- c$ and multiply signs pair by pair.
Usually costs: 1 to 2 marks.
Mistake: Choosing a pair with the right product but wrong sum
Students stop at the first pair of factors they find.
Wrong working
$x^{2} + 11 x + 30 = (x + 3) (x + 10)$
Right working
$3 + 10 = 13 \neq = 11$ ; the pair is 5 and 6: $(x + 5) (x + 6)$
Do this: List factor pairs of the product and pick the one with the right sum; expand to check.
Usually costs: The whole factorisation.
Mistake: Cancelling terms instead of factors
The fraction looks like it has the same letters on top and bottom.
Wrong working
$\frac{x ^{2} + 3}{x ^{2} + 5} = \frac{3}{5}$
Right working
Neither $x^{2} + 3$ nor $x^{2} + 5$ factorises further and they share no factor, so the expression is already in simplest form.
Do this: Factorise first; cancel only whole brackets that appear as factors in both.
Usually costs: All marks for the simplification.
Mistake: Forgetting the common factor
Students try to match an identity before taking out a common number.
Wrong working
$3 x^{2} - 27$ cannot be factorised because 3 and 27 are not squares.
Right working
$3 x^{2} - 27 = 3 (x^{2} - 9) = 3 (x + 3) (x - 3)$
Do this: Always look for a common factor first.
Usually costs: 1 to 2 marks.

NCERT exercise guide: Exploring Algebraic Identities

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise Set 4.1

What it tests: Expanding squares of binomials with whole-number, decimal and fractional coefficients, and squaring numbers such as 64, 105 and 205 with $(a + b)^{2}$ .

How to approach it: Identify a and b with their coefficients, square each, add twice the product. For numbers, split into a round part plus a small part.

Like the expansion and number-squaring questions in Exercise Set 4.1

Expand $(\frac{x}{2} + \frac{2 y}{3})^{2}$ and find $10 7^{2}$ .

Show the solutionSolution

$(\frac{x}{2})^{2} + 2 \times \frac{x}{2} \times \frac{2 y}{3} + (\frac{2 y}{3})^{2} = \frac{x ^{2}}{4} + \frac{2 x y}{3} + \frac{4 y ^{2}}{9}$ .
$10 7^{2} = (100 + 7)^{2} = 10000 + 1400 + 49 = 11449$ .

Answer: $\frac{x ^{2}}{4} + \frac{2 x y}{3} + \frac{4 y ^{2}}{9}$ ; $11449$

Exercise Set 4.2

What it tests: Factorising perfect squares (some after taking out a common factor) and squaring numbers such as 79 and 299 with $(a - b)^{2}$ .

How to approach it: Check the end terms are squares and the middle term is twice their product. Take out a common factor first when the coefficients are not squares.

Like the factorising questions in Exercise Set 4.2

Factorise $18 m^{2} + 48 mn + 32 n^{2}$ and find $18 9^{2}$ .

Show the solutionSolution

Common factor 2: $2 (9 m^{2} + 24 mn + 16 n^{2})$ .
$9 m^{2} = (3 m)^{2}$ , $16 n^{2} = (4 n)^{2}$ , $2 (3 m) (4 n) = 24 mn$ , so $2 (3 m + 4 n)^{2}$ .
$18 9^{2} = (200 - 11)^{2} = 40000 - 4400 + 121 = 35721$ .

Answer: $2 (3 m + 4 n)^{2}$ ; $35721$

Exercise Set 4.3

What it tests: Squaring numbers with the most convenient identity, factorising with $(a + b + c)^{2}$ , expanding trinomial squares, and testing whether a given equality is an identity.

How to approach it: For six-term expressions, take square roots of the three square terms and fix the signs so every product term matches. To test an identity, expand both sides fully.

Like the trinomial factorising questions in Exercise Set 4.3

Factorise $4 a^{2} + 9 b^{2} + c^{2} - 12 ab + 6 b c - 4 a c$ .

Show the solutionSolution

Square roots of the square terms: $2 a$ , $3 b$ , $c$ (signs to be fixed).
$- 12 ab$ needs a and b of opposite signs; $- 4 a c$ needs a and c of opposite signs; $+ 6 b c$ needs b and c of the same sign.
So take $2 a$ , $- 3 b$ , $- c$ : check $2 (- 3 b) (- c) = 6 b c$ .
The expression is $(2 a - 3 b - c)^{2}$ .

Answer: $(2 a - 3 b - c)^{2}$

Exercise Set 4.4

What it tests: Completing factorisations of quadratics, choosing identities for products like $23 \times 17$ and $9 7^{2}$ , and factorising trinomial squares and quadratics.

How to approach it: For $a x^{2} + b x + c$ , find two numbers with product ac and sum b; split and group. For number products, look for a common midpoint.

Like the quadratic factorising questions in Exercise Set 4.4

Factorise $6 x^{2} - 7 x - 20$ and find $46 \times 54$ .

Show the solutionSolution

$a c = - 120$ , $b = - 7$ : the numbers are $- 15$ and 8.
$6 x^{2} - 15 x + 8 x - 20 = 3 x (2 x - 5) + 4 (2 x - 5) = (3 x + 4) (2 x - 5)$ .
$46 \times 54 = (50 - 4) (50 + 4) = 2500 - 16 = 2484$ .

Answer: $(3 x + 4) (2 x - 5)$ ; $2484$

Exercise Set 4.5

What it tests: Simplifying rational expressions in one or two variables by factorising, assuming denominators are not zero.

How to approach it: Factorise numerator and denominator completely (common factor, identity or splitting), cancel common factors, and write the result.

Like the rational expressions in Exercise Set 4.5

Simplify $\frac{2 x ^{2} + 5 x - 3}{4 x ^{2} - 1}$ , assuming the denominator is not zero.

Show the solutionSolution

Numerator: $a c = - 6$ , $b = 5$ : numbers 6 and $- 1$ . $2 x^{2} + 6 x - x - 3 = (2 x - 1) (x + 3)$ .
Denominator: $(2 x - 1) (2 x + 1)$ .
Cancel $(2 x - 1)$ : $\frac{x + 3}{2 x + 1}$ .

Answer: $\frac{x + 3}{2 x + 1}$

End-of-chapter Exercises

What it tests: Products by identities, values like $19 9^{3}$ and $104 \times 96$ , factorising many types of expressions, simplifying, finding sides of rectangles and cuboids from areas and volumes, and word problems that lead to quadratics. Starred questions are enrichment.

How to approach it: Decide which identity fits by counting terms and spotting squares or cubes. For area or volume questions, factorise completely; each factor is a length.

Like the cube-value and cuboid questions at the end of the chapter

Find $20 1^{3}$ using an identity, and find possible dimensions of a cuboid whose volume is $3 p s^{2} - 12 p s + 9 p$ .

Show the solutionSolution

$20 1^{3} = (200 + 1)^{3} = 8000000 + 3 (40000) (1) + 3 (200) (1) + 1 = 8120601$ .
$3 p s^{2} - 12 p s + 9 p = 3 p (s^{2} - 4 s + 3)$ .
$s^{2} - 4 s + 3 = (s - 1) (s - 3)$ (sum $- 4$ , product 3).
Volume $= 3 p (s - 1) (s - 3)$ .

Answer: $8120601$ ; dimensions $3 p$ , $s - 1$ and $s - 3$ .

Practice questions on Exploring Algebraic Identities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The coefficient of x in $(x - 4) (x + 9)$ is
(a) $- 36$
(b) 5
(c) $- 5$
(d) 13
Hint
Use $(x + a) (x + b)$ with $a = - 4$ , $b = 9$ .
Answer
(b) 5
Q2 · 1 mark · MCQExam level
The value of $4 9^{2} - 5 1^{2}$ is
(a) $- 200$
(b) 200
(c) $- 100$
(d) 4
Hint
$a^{2} - b^{2} = (a + b) (a - b)$ .
Answer
(a) $- 200$ , since $(49 + 51) (49 - 51) = 100 \times (- 2)$ .
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): $x^{2} + 6 x + 9 = (x + 3)^{2}$ for every value of x.
Reason (R): An identity is true for all values of its variables.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is A an instance of $(a + b)^{2}$ ?
Answer
(a). A is the identity $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ with $a = x$ , $b = 3$ , and R explains why it holds for every x.
Q4 · 2 marks · Very short answerExam level
Expand $(2 x + 3 y + z)^{2}$ .
Hint
Three squares and three doubled products.
Answer
$4 x^{2} + 9 y^{2} + z^{2} + 12 x y + 6 y z + 4 z x$
Q5 · 2 marks · Very short answerHarder exam level
Factorise $27 a^{3} - 8 b^{3}$ .
Hint
Difference of cubes with $x = 3 a$ , $y = 2 b$ .
Answer
$(3 a - 2 b) (9 a^{2} + 6 ab + 4 b^{2})$
Q6 · 2 marks · Very short answerHarder exam level
Evaluate $10 3^{3}$ using an identity.
Hint
$(100 + 3)^{3}$ .
Answer
$1000000 + 90000 + 2700 + 27 = 1092727$
Q7 · 3 marks · Short answerHarder exam level
Factorise (i) $x^{2} - 13 x + 42$ (ii) $3 x^{2} + 10 x + 8$ .
Hint
(ii): product 24, sum 10.
Answer
(i) $(x - 6) (x - 7)$ (ii) $3 x^{2} + 6 x + 4 x + 8 = (3 x + 4) (x + 2)$
Q8 · 3 marks · Short answerHarder exam level
If $a - b = 4$ and $ab = 21$ , find $a^{2} + b^{2}$ and $a^{3} - b^{3}$ .
Hint
Use $(a - b)^{2}$ and $(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)$ .
Answer
$a^{2} + b^{2} = 16 + 42 = 58$ ; $a^{3} - b^{3} = 64 + 3 (21) (4) = 316$ .
Q9 · 3 marks · Short answerHarder exam level
Simplify $\frac{p ^{2} - 16}{p ^{2} + 2 p - 8}$ , assuming the denominator is not zero.
Hint
Factorise both.
Answer
$\frac{( p - 4 ) ( p + 4 )}{( p + 4 ) ( p - 2 )} = \frac{p - 4}{p - 2}$
Q10 · 5 marks · Long answerHarder exam level
A rectangle has area $6 x^{2} + 13 x + 6$ square units. Find expressions for its sides and its perimeter. Find the actual sides and area when $x = 2$ , and check the area from the expression.
Hint
Product 36, sum 13.
Answer
$6 x^{2} + 9 x + 4 x + 6 = (2 x + 3) (3 x + 2)$ ; perimeter $= 10 x + 10$ . At $x = 2$ : sides 7 and 8, area 56; expression: $24 + 26 + 6 = 56$ .
Q11 · 3 marks · Short answerHOTS / competency
Show that for any four consecutive square numbers $n^{2}$ , $(n + 1)^{2}$ , $(n + 2)^{2}$ , $(n + 3)^{2}$ , the first minus the second minus the third plus the fourth is always 4. Test it with 4, 9, 16, 25.
Hint
Expand each square.
Answer
$n^{2} - (n^{2} + 2 n + 1) - (n^{2} + 4 n + 4) + (n^{2} + 6 n + 9) = 4$ . Test: $4 - 9 - 16 + 25 = 4$ .
Q12 · 3 marks · Short answerHOTS / competency
Without finding the cubes, evaluate $(- 15)^{3} + 8^{3} + 7^{3}$ . Explain the identity you use.
Hint
Add the three numbers first.
Answer
$- 15 + 8 + 7 = 0$ , so the sum of cubes $= 3 x y z = 3 (- 15) (8) (7) = - 2520$ , from $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (\dots)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the difference between an identity and an equation?Show answer
An identity is true for all values of the variables; an equation is true only for some.
2.Write $(a - b)^{2}$ .Show answer
$a^{2} - 2 ab + b^{2}$
3.Write $(a + b + c)^{2}$ .Show answer
$a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$
4.How did Śrīdharāchārya square numbers?Show answer
Using $a^{2} = (a + b) (a - b) + b^{2}$ , e.g. $7 5^{2} = 80 \times 70 + 25$ .
5.Expand $(x + a) (x + b)$ .Show answer
$x^{2} + (a + b) x + ab$
6.Write $(a - b)^{3}$ .Show answer
$a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$
7.Factorise $x^{3} + y^{3}$ .Show answer
$(x + y) (x^{2} - x y + y^{2})$
8.If $x + y + z = 0$ , what is $x^{3} + y^{3} + z^{3}$ ?Show answer
$3 x y z$
9.What can you cancel in a rational expression?Show answer
Only common factors of the numerator and denominator, assuming they are not zero.
10.Why do three consecutive squares give $(n - 1)^{2} + (n + 1)^{2} - 2 n^{2} = 2$ ?Show answer
Expanding gives $2 n^{2} + 2 - 2 n^{2} = 2$ for every n.

Best order to study it

Learn the difference between an identity and an equation, and draw the $(a + b)^{2}$ square model yourself.
Practise $(a \pm b)^{2}$ expansions and number squares (Exercise Set 4.1), then perfect-square factorising (Set 4.2).
Add $(a + b + c)^{2}$ and $a^{2} - b^{2}$ , including Śrīdharāchārya's method (Set 4.3).
Learn splitting the middle term with algebra tiles first, then without them (Set 4.4).
Learn the cube identities and the three-cube identity; practise value questions.
Finish with rational expressions (Set 4.5) and the mixed end-of-chapter exercises.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw the square model for $(a + b)^{2}$ and the cube model for $(a + b)^{3}$ ?
Can you show by a counter-example that $(a + b)^{2} = a^{2} + b^{2}$ is not an identity?
Can you find $9 7^{2}$ , $10 5^{2}$ and $46 \times 54$ mentally?
Can you factorise $25 a^{2} - 30 ab + 9 b^{2}$ ?
Can you expand $(p - 2 q + 3 r)^{2}$ with all signs right?
Can you factorise $3 x^{2} - x - 10$ by splitting the middle term?
Can you factorise $64 y^{3} + 1$ ?
Can you find $a^{3} + b^{3} + c^{3}$ when $a + b + c = 0$ ?
Can you find $x^{2} + \frac{1}{x ^{2}}$ given $x + \frac{1}{x} = 4$ ?
Can you simplify $\frac{x ^{2} - 16}{x ^{2} - x - 12}$ and state the excluded values?

Exploring Algebraic Identities: frequently asked questions

Which identities must I learn for Class 9 (2026-27)?

The NCERT chapter summary lists eleven: (x \pm y)^2, (x + y + z)^2, (x + y)(x - y), (x + a)(x + b), (ax + b)(cx + d), x^3 \pm y^3, (x \pm y)^3 and x^3 + y^3 + z^3 - 3xyz. Learn all of them with one example each.

Do I need to draw algebra tiles in the exam?

The CBSE syllabus includes visualising factorisation with algebra tiles, so you may be asked to show or explain a tile arrangement, such as the rectangle for x^2 + 7x + 12. Practise drawing one or two neatly.

Is factorising by the factor theorem part of this chapter?

No. In the 2026-27 syllabus, quadratics are factorised by identities and splitting the middle term. The factor theorem is not listed.

How do I know which identity to use?

Count the terms and look at the powers: two square terms with a minus between them suggest a^2 - b^2; three terms with squares at both ends suggest a perfect square; two cube terms suggest a sum or difference of cubes; four terms with cubes at both ends suggest (a \pm b)^3.

Why do I have to say the denominator is not zero when simplifying?

Cancelling a factor means dividing by it, which is only allowed when it is not zero. The simplified expression is equal to the original only where the original is defined.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part I, Mathematics Textbook for Grade 9 (First Edition 2026)