MathsICSE Class 9 · Chapter 3

Expansions ICSE Class 9: Algebraic Identities with Solved Questions

Updated 15 October 2026 · Fact-checked

Expansion means multiplying out brackets using standard identities. ICSE Class 9 uses four: (a ± b)², (a ± b)³, (x ± a)(x ± b) and (a ± b ± c)². You expand expressions, find values such as a² + b² or x³ + 1/x³ from given sums and products, and use identities to calculate squares and cubes of numbers quickly.

What Expansions is about

An identity is an equation that is true for every value of its variables, such as $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ . Expanding means using such identities to write a product of brackets as a sum of terms.

The CISCE 2028 Class IX syllabus lists the expansions $(a \pm b)^{2}$ , $(a \pm b)^{3}$ , $(x \pm a) (x \pm b)$ and $(a \pm b \pm c)^{2}$ , after recalling what you learnt in earlier classes. School examiners use them in three ways: straight expansion, finding the value of an expression from given data (for example $a^{2} + b^{2}$ when $a + b$ and $ab$ are known, or $x^{3} + \frac{1}{x ^{3}}$ when $x + \frac{1}{x}$ is known), and quick arithmetic such as $10 3^{2}$ or $9 9^{3}$ .

The same identities run backwards in the next chapter, Factorisation, so mastering them here pays twice.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Expansions is in the 3. Algebra (i) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Square of a binomial: (a ± b)²

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

The middle term is twice the product of the two terms. Treat each term as a whole, including its coefficient and sign: in $(3 x - 2 y)^{2}$ , $a = 3 x$ and $b = 2 y$ .

Useful rearrangements:

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

Worked example

Expand $(2 x - \frac{1}{3 x})^{2}$ .

Try it first, then show the solutionSolution

$a = 2 x$ , $b = \frac{1}{3 x}$ .
$a^{2} = 4 x^{2}$ , $2 ab = 2 \times 2 x \times \frac{1}{3 x} = \frac{4}{3}$ , $b^{2} = \frac{1}{9 x ^{2}}$ .
$(2 x - \frac{1}{3 x})^{2} = 4 x^{2} - \frac{4}{3} + \frac{1}{9 x ^{2}}$ .

Answer: $4 x^{2} - \frac{4}{3} + \frac{1}{9 x ^{2}}$

Product of the form (x + a)(x + b)

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

The coefficient of $x$ is the sum of the two numbers and the constant is their product, with signs. So $(x + 5) (x - 3) = x^{2} + 2 x - 15$ and $(x - 4) (x - 6) = x^{2} - 10 x + 24$ .

The identity also works when the first terms match but are not just $x$ : $(2 y + 3) (2 y - 7) = 4 y^{2} - 8 y - 21$ .

Worked example

Use $(x + a) (x + b)$ to find $103 \times 104$ .

Try it first, then show the solutionSolution

$103 \times 104 = (100 + 3) (100 + 4)$ .
$= 10 0^{2} + (3 + 4) \times 100 + 12 = 10000 + 700 + 12$ .
$= 10712$ .

Answer: 10712

Cube of a binomial: (a ± b)³

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

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

The coefficients 1, 3, 3, 1 come from multiplying $(a + b)^{2}$ by $(a + b)$ . In $(a - b)^{3}$ the signs alternate $+, -, +, -$ .

Rearranged forms used for 'find the value' questions:

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

Worked example

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

Try it first, then show the solutionSolution

$a = 3 x$ , $b = 2$ .
$a^{3} = 27 x^{3}$ , $3 a^{2} b = 3 \times 9 x^{2} \times 2 = 54 x^{2}$ , $3 a b^{2} = 3 \times 3 x \times 4 = 36 x$ , $b^{3} = 8$ .
Signs $+, -, +, -$ : $27 x^{3} - 54 x^{2} + 36 x - 8$ .

Answer: $27 x^{3} - 54 x^{2} + 36 x - 8$

Square of a trinomial: (a ± b ± c)²

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

Square each term, then add twice the product of every pair. For other sign patterns, put the sign inside the term: $(a - b + c)^{2}$ uses $a$ , $- b$ , $c$ , so it equals $a^{2} + b^{2} + c^{2} - 2 ab - 2 b c + 2 c a$ .

Rearranged: $a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a)$ .

Worked example

Expand $(a + 2 b - c)^{2}$ .

Try it first, then show the solutionSolution

Terms: $a$ , $2 b$ , $- c$ .
Squares: $a^{2} + 4 b^{2} + c^{2}$ .
Pairs: $2 (a) (2 b) = 4 ab$ , $2 (2 b) (- c) = - 4 b c$ , $2 (- c) (a) = - 2 c a$ .
$a^{2} + 4 b^{2} + c^{2} + 4 ab - 4 b c - 2 c a$ .

Answer: $a^{2} + 4 b^{2} + c^{2} + 4 ab - 4 b c - 2 c a$

Finding values from sums and products

Many questions give two facts and ask for a third. Choose the identity that links them.

Given $a + b$ and $ab$ : $a^{2} + b^{2} = (a + b)^{2} - 2 ab$ ; $(a - b)^{2} = (a + b)^{2} - 4 ab$ ; $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ .
Given $a - b$ and $ab$ : $a^{3} - b^{3} = (a - b)^{3} + 3 ab (a - b)$ .
Given $a + b + c$ and $ab + b c + c a$ : $a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a)$ .

When you take a square root, remember both signs unless the question says the numbers are positive.

Worked example

If $a + b = 7$ and $ab = 10$ , find (i) $a^{2} + b^{2}$ (ii) $a - b$ (iii) $a^{3} + b^{3}$ .

Try it first, then show the solutionSolution

(i) $49 - 20 = 29$ .
(ii) $(a - b)^{2} = 49 - 40 = 9$ , so $a - b = \pm 3$ .
(iii) $343 - 3 \times 10 \times 7 = 343 - 210 = 133$ .

Answer: (i) 29 (ii) ±3 (iii) 133

x + 1/x type problems

Put $a = x$ and $b = \frac{1}{x}$ , so $ab = 1$ . Then:

$x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2 = (x - \frac{1}{x})^{2} + 2$
$x^{3} + \frac{1}{x ^{3}} = (x + \frac{1}{x})^{3} - 3 (x + \frac{1}{x})$
$x^{3} - \frac{1}{x ^{3}} = (x - \frac{1}{x})^{3} + 3 (x - \frac{1}{x})$
$x^{4} + \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}})^{2} - 2$

To move from $x - \frac{1}{x}$ to $x + \frac{1}{x}$ , go through $x^{2} + \frac{1}{x ^{2}}$ : $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x ^{2}} + 2$ .

Worked example

If $x + \frac{1}{x} = 4$ , find $x^{2} + \frac{1}{x ^{2}}$ and $x^{3} + \frac{1}{x ^{3}}$ .

Try it first, then show the solutionSolution

$x^{2} + \frac{1}{x ^{2}} = 16 - 2 = 14$ .
$x^{3} + \frac{1}{x ^{3}} = 64 - 3 \times 4 = 52$ .

Answer: 14 and 52

Maths in real life: Expansions

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 as an area

Suppose a square garden of side $x$ metres is widened by 2 m on one side and lengthened by 2 m on the other, so it stays square.

Find the increase in area in terms of $x$ , and its value when $x = 10$ .

Try it first, then show the solutionSolution

New area $= (x + 2)^{2} = x^{2} + 4 x + 4$ .
Increase $= 4 x + 4$ ; at $x = 10$ it is 44 m².

Answer: (4x + 4) m², i.e. 44 m² when x = 10.

Memory hook: The 2ab strips plus the small corner square b².

Quick squares with (a − b)²

Imagine a shopkeeper pricing 98 boxes of 98 pencils each and wanting the total without a calculator.

Find $9 8^{2}$ mentally.

Try it first, then show the solutionSolution

$9 8^{2} = (100 - 2)^{2} = 10000 - 400 + 4$ .
$= 9604$ .

Answer: 9,604 pencils.

Memory hook: Near 100? Use (100 ± b)².

Cube of a sum as volume

Suppose a cube-shaped water tank has edge 1 m, and a design change adds 2 cm to every edge.

Find the new volume in cubic metres using $(a + b)^{3}$ .

Try it first, then show the solutionSolution

$(1 + 0.02)^{3} = 1 + 3 (0.02) + 3 (0.02)^{2} + (0.02)^{3}$ .
$= 1 + 0.06 + 0.0012 + 0.000008 = 1.061208$ .

Answer: 1.061208 m³ (about 6.1% more).

Memory hook: Small change b: the volume grows by about 3b.

Topics in this chapter

Expansions formula sheet

Square of a sum: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$; Middle term: twice the product.
Square of a difference: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$; Only the middle sign changes.
Common first term: $(x + a) (x + b) = x^{2} + (a + b) x + ab$; Sum and product of a and b.
Cube of a sum: $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$; Long form: a³ + 3a²b + 3ab² + b³.
Cube of a difference: $(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)$; Signs alternate: +, −, +, −.
Square of a trinomial: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$; Put signs inside the terms.
Reciprocal pair: $x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2$; Use + 2 with x − 1/x.

How the school exam asks Expansions

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

Question type	How it is asked	Method
Direct expansion	Expand (2a − 3b)², (x + 2y)³, (a − b + 2c)² or (x − 7)(x + 4).	Identify a, b (and c) with their signs and coefficients, then apply the identity term by term.
Find the value from given data	Given a + b and ab (or a − b and a² + b²), find a² + b², a³ + b³, a − b and so on.	Pick the rearranged identity that contains both the given and the required quantities.
Reciprocal (x ± 1/x) problems	Given x + 1/x or x² + 1/x², find higher powers.	Use ab = 1 in the identities; move step by step through squares.
Quick arithmetic	Evaluate 103², 998², 99³ or 1.02³ using identities.	Write the number as (round number ± small number) and expand.
Simplify an expression	Simplify (x + 2y)² − (x − 2y)², or (a + b)³ − (a − b)³.	Expand both or use (a + b)² − (a − b)² = 4ab; collect like terms carefully.
Three-variable values	Given a + b + c and a² + b² + c², find ab + bc + ca.	(a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Tricks to remember Expansions

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

First, twice, last
(first ± last)² = first² ± twice first × last + last².
Use it when: Binomial squares.
1-3-3-1
Cube coefficients are 1, 3, 3, 1, and the power of a falls 3, 2, 1, 0 while b rises 0, 1, 2, 3.
Use it when: Writing (a ± b)³ in full.
Minus cube: plus-minus-plus-minus
(a − b)³ signs go + − + −, like a swing.
Use it when: Avoiding sign slips in cubes.
Every pair, twice
Trinomial square: three squares plus every pair doubled (ab, bc, ca).
Use it when: (a ± b ± c)².

Expansions: exam-level solved questions

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

Example 1: Coefficient in an expansion · 1 mark · MCQExam level

The coefficient of $x$ in $(x - 7) (x + 4)$ is:
(a) 3
(b) $- 3$
(c) $- 28$
(d) 11

Try it first, then show the solutionSolution

Coefficient of $x$ $= - 7 + 4 = - 3$ .

Answer: (b)

Example 2: Trinomial square · 3 marks · Short answerExam level

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

Try it first, then show the solutionSolution

Terms: $2 x$ , $- 3 y$ , $z$ .
Squares: $4 x^{2} + 9 y^{2} + z^{2}$ .
Pairs: $2 (2 x) (- 3 y) = - 12 x y$ , $2 (- 3 y) (z) = - 6 y z$ , $2 (z) (2 x) = 4 z x$ .
$4 x^{2} + 9 y^{2} + z^{2} - 12 x y - 6 y z + 4 z x$ .

Answer: $4 x^{2} + 9 y^{2} + z^{2} - 12 x y - 6 y z + 4 z x$

Example 3: Evaluate using identities · 3 marks · Short answerExam level

Using suitable identities, evaluate (i) $99 7^{2}$ (ii) $9 9^{3}$ .

Try it first, then show the solutionSolution

(i) $(1000 - 3)^{2} = 1000000 - 6000 + 9 = 994009$ .
(ii) $(100 - 1)^{3} = 1000000 - 1 - 3 \times 100 \times 1 \times 99$ .
$= 999999 - 29700 = 970299$ .

Answer: (i) 994009 (ii) 970299

Example 4: From a difference to a cube · 3 marks · Short answerHarder exam level

If $a - b = 4$ and $a^{2} + b^{2} = 40$ , find $ab$ and $a^{3} - b^{3}$ .

Try it first, then show the solutionSolution

$(a - b)^{2} = a^{2} + b^{2} - 2 ab$ : $16 = 40 - 2 ab$ , so $ab = 12$ .
$a^{3} - b^{3} = (a - b)^{3} + 3 ab (a - b) = 64 + 3 \times 12 \times 4$ .
$= 64 + 144 = 208$ .

Answer: ab = 12; a³ − b³ = 208

Example 5: Sum of three squares · 3 marks · Short answerHarder exam level

If $a + b + c = 12$ and $a^{2} + b^{2} + c^{2} = 50$ , find $ab + b c + c a$ .

Try it first, then show the solutionSolution

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$ .
$144 = 50 + 2 (ab + b c + c a)$ .
$ab + b c + c a = 47$ .

Answer: 47

Example 6: Cubes with coefficients · 4 marks · Long answerHarder exam level

If $2 x + 3 y = 13$ and $x y = 6$ , find the value of $8 x^{3} + 27 y^{3}$ .

Try it first, then show the solutionSolution

Let $a = 2 x$ , $b = 3 y$ : $a + b = 13$ , $ab = 6 x y = 36$ .
$a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ .
$= 2197 - 3 \times 36 \times 13 = 2197 - 1404 = 793$ .

Answer: 793

Example 7: Reciprocal chain from x − 1/x · 4 marks · Long answerHOTS / competency

If $x - \frac{1}{x} = 3$ , find (i) $x^{2} + \frac{1}{x ^{2}}$ (ii) $x^{3} - \frac{1}{x ^{3}}$ (iii) $x + \frac{1}{x}$ , given $x > 0$ .

Try it first, then show the solutionSolution

(i) $x^{2} + \frac{1}{x ^{2}} = 9 + 2 = 11$ .
(ii) $x^{3} - \frac{1}{x ^{3}} = 27 + 3 \times 3 = 36$ .
(iii) $(x + \frac{1}{x})^{2} = 11 + 2 = 13$ ; since $x > 0$ , $x + \frac{1}{x} = 13$ .

Answer: (i) 11 (ii) 36 (iii) √13

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

A tiled courtyard

A square courtyard has side $(a + b)$ metres. It is tiled with one square of side $a$ , one square of side $b$ and two rectangles of size $a \times b$ . Later the owners build a cube-shaped storage box of edge $(a + b)$ dm. For their courtyard, $a + b = 9$ and $ab = 20$ .

(i) · 3 marks
Using the tiles, explain why $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , and find the total area of the two square tiles.
Answer
Answer: 41 m²
The whole square is made of the four pieces, so $(a + b)^{2} = a^{2} + ab + ab + b^{2}$ . Square tiles: $a^{2} + b^{2} = 81 - 40 = 41$ m².
(ii) · 3 marks
Find $a - b$ , given $a > b$ , and hence the side of each square tile.
Answer
Answer: a − b = 1; a = 5, b = 4
$(a - b)^{2} = 81 - 80 = 1$ , so $a - b = 1$ . With $a + b = 9$ : $a = 5$ , $b = 4$ .
(iii) · 4 marks
The box's volume is $(a + b)^{3}$ dm³. Find $a^{3} + b^{3}$ using an identity and check with $a$ and $b$ .
Answer
Answer: 189
$a^{3} + b^{3} = 729 - 3 \times 20 \times 9 = 729 - 540 = 189$ . Check: $125 + 64 = 189$ .

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 $(2 a - 3 b)^{3}$ . (3 marks)

Step	Marks
Identify $a \to 2 a$ , $b \to 3 b$ in $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$	1
Terms: $8 a^{3}$ , $3 (4 a^{2}) (3 b) = 36 a^{2} b$ , $3 (2 a) (9 b^{2}) = 54 a b^{2}$ , $27 b^{3}$	1
$8 a^{3} - 36 a^{2} b + 54 a b^{2} - 27 b^{3}$	1

If $x + \frac{1}{x} = 5$ , find $x^{2} + \frac{1}{x ^{2}}$ and $x^{4} + \frac{1}{x ^{4}}$ . (4 marks)

Step	Marks
Square: $x^{2} + 2 + \frac{1}{x ^{2}} = 25$	1
$x^{2} + \frac{1}{x ^{2}} = 23$	1
Square again: $x^{4} + 2 + \frac{1}{x ^{4}} = 529$	1
$x^{4} + \frac{1}{x ^{4}} = 527$	1

If $x - \frac{1}{x} = 4$ and $x > 0$ , find $x^{2} + \frac{1}{x ^{2}}$ , $x + \frac{1}{x}$ , $x^{2} - \frac{1}{x ^{2}}$ and $x^{3} - \frac{1}{x ^{3}}$ . (5 marks)

Step	Marks
$x^{2} + \frac{1}{x ^{2}} = 16 + 2 = 18$	1
$(x + \frac{1}{x})^{2} = 18 + 2 = 20$	1
$x + \frac{1}{x} = 25$ (positive since $x > 0$ )	1
$x^{2} - \frac{1}{x ^{2}} = (x + \frac{1}{x}) (x - \frac{1}{x}) = 85$	1
$x^{3} - \frac{1}{x ^{3}} = 64 + 3 \times 4 = 76$	1

Where students lose marks in Expansions

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

Mistake: Dropping the middle term
Writing $(a + b)^{2} = a^{2} + b^{2}$ is the most common expansion error.
Wrong working
$(x + 3)^{2} = x^{2} + 9$
Right working
$(x + 3)^{2} = x^{2} + 6 x + 9$
Do this: Say 'first, twice, last' as you write.
Usually costs: The whole expansion.
Mistake: Squaring only the variable, not the coefficient
In $(3 x)^{2}$ students write $3 x^{2}$ instead of $9 x^{2}$ .
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 each term in brackets before squaring or cubing.
Usually costs: 1 mark per term.
Mistake: Sign errors in (a − b)³ and trinomial squares
Pair products with one negative term must be negative; students write all pair terms as positive.
Do this: Write each term with its sign ( $a$ , $- b$ , $c$ ) and multiply signs pair by pair.
Usually costs: 1 to 2 marks.
Mistake: Taking only the positive square root
From $(a - b)^{2} = 9$ , students write $a - b = 3$ when $\pm 3$ is correct unless the question says $a > b$ .
Do this: Write $\pm$ and then use any condition given to choose.
Usually costs: 1 mark.
Mistake: Using + 2 instead of − 2 (or the reverse)
$x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2$ but $= (x - \frac{1}{x})^{2} + 2$ ; students mix them.
Do this: Expand $(x \pm \frac{1}{x})^{2}$ in full each time; the middle term $\pm 2$ shows you which way to go.
Usually costs: Every later part.

Practice questions on Expansions

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
$(a + b)^{2} - (a - b)^{2}$ equals:
(a) $2 ab$
(b) $4 ab$
(c) $2 (a^{2} + b^{2})$
(d) 0
Hint
Expand both.
Answer
(b)
Q2 · 2 marks · Short answerExam level
Expand $(3 x + 2 y)^{2}$ and $(y - 8) (y + 5)$ .
Hint
Twice the product; sum and product.
Answer
$9 x^{2} + 12 x y + 4 y^{2}$ ; $y^{2} - 3 y - 40$ .
Q3 · 2 marks · Short answerExam level
If $a + b = 9$ and $a - b = 5$ , find $ab$ .
Hint
(a + b)² − (a − b)² = 4ab.
Answer
$4 ab = 81 - 25 = 56$ , so $ab = 14$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $(x - y)^{3} = x^{3} - y^{3}$ for all $x$ and $y$ .
Reason (R): $(x - y)^{3} = x^{3} - y^{3} - 3 x y (x - y)$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Try x = 2, y = 1.
Answer
(b). $(2 - 1)^{3} = 1$ but $8 - 1 = 7$ ; (R) is the correct identity.
Q5 · 3 marks · Short answerHarder exam level
Expand $(x + \frac{2}{x})^{3}$ .
Hint
a = x, b = 2/x.
Answer
$x^{3} + 3 x^{2} \cdot \frac{2}{x} + 3 x \cdot \frac{4}{x ^{2}} + \frac{8}{x ^{3}} = x^{3} + 6 x + \frac{12}{x} + \frac{8}{x ^{3}}$ .
Q6 · 3 marks · Short answerHarder exam level
Simplify $(2 x + 3)^{2} - (2 x - 3)^{2} + (x + 1) (x - 1)$ .
Hint
Use 4ab for the first two terms.
Answer
$4 \times 2 x \times 3 + x^{2} - 1 = x^{2} + 24 x - 1$ .
Q7 · 4 marks · Short answerHarder exam level
If $x + \frac{1}{x} = 7$ , find $x^{2} + \frac{1}{x ^{2}}$ , $x^{3} + \frac{1}{x ^{3}}$ and $x - \frac{1}{x}$ (given $x > 1$ ).
Hint
Square, then cube.
Answer
$49 - 2 = 47$ ; $343 - 21 = 322$ ; $(x - \frac{1}{x})^{2} = 47 - 2 = 45$ , so $x - \frac{1}{x} = 35$ (positive since $x > 1$ ).
Q8 · 4 marks · Long answerHarder exam level
If $a + b + c = 10$ and $ab + b c + c a = 31$ , find $a^{2} + b^{2} + c^{2}$ . Hence find $(a - b)^{2} + (b - c)^{2} + (c - a)^{2}$ .
Hint
The second sum equals 2(a² + b² + c²) − 2(ab + bc + ca).
Answer
$a^{2} + b^{2} + c^{2} = 100 - 62 = 38$ . Second: $2 \times 38 - 2 \times 31 = 14$ .
Q9 · 3 marks · Short answerHarder exam level
Find the product $(x + 1) (x + 2) (x + 3)$ using $(x + a) (x + b)$ twice.
Hint
Multiply the first two, then by (x + 3).
Answer
$(x^{2} + 3 x + 2) (x + 3) = x^{3} + 6 x^{2} + 11 x + 6$ .
Q10 · 4 marks · Long answerHarder exam level
If $x^{2} + \frac{1}{x ^{2}} = 23$ and $x > 0$ , find $x + \frac{1}{x}$ and $x^{3} + \frac{1}{x ^{3}}$ .
Hint
Add 2 and take the square root.
Answer
$(x + \frac{1}{x})^{2} = 25$ , so $x + \frac{1}{x} = 5$ ; $x^{3} + \frac{1}{x ^{3}} = 125 - 15 = 110$ .
Q11 · 4 marks · Long answerHOTS / competency
If $x = 3 + 22$ , find $x^{2} + \frac{1}{x ^{2}}$ and $x^{3} + \frac{1}{x ^{3}}$ .
Hint
First find 1/x by rationalising.
Answer
$\frac{1}{x} = 3 - 22$ , so $x + \frac{1}{x} = 6$ . Then $x^{2} + \frac{1}{x ^{2}} = 34$ and $x^{3} + \frac{1}{x ^{3}} = 216 - 18 = 198$ .
Q12 · 4 marks · Long answerHOTS / competency
Without multiplying out, show that $(a + b)^{3} - (a - b)^{3} = 2 b (3 a^{2} + b^{2})$ and use it to find $1 3^{3} - 7^{3}$ .
Hint
Subtract the two long forms; then take a = 10, b = 3.
Answer
$(a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}) - (a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}) = 6 a^{2} b + 2 b^{3} = 2 b (3 a^{2} + b^{2})$ . With $a = 10$ , $b = 3$ : $6 (300 + 9) = 1854$ . Check: $2197 - 343 = 1854$ .

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.Expand (a − b)².Show answer
a² − 2ab + b².
2.Expand (x + a)(x + b).Show answer
x² + (a + b)x + ab.
3.Write (a + b)³ in short form.Show answer
a³ + b³ + 3ab(a + b).
4.What is (a + b)² − (a − b)²?Show answer
4ab.
5.Express a² + b² + c² in terms of a + b + c and ab + bc + ca.Show answer
(a + b + c)² − 2(ab + bc + ca).
6.If x + 1/x = 3, what is x² + 1/x²?Show answer
7.
7.What are the coefficients in (a + b)³?Show answer
1, 3, 3, 1.
8.What is x³ − 1/x³ in terms of x − 1/x?Show answer
(x − 1/x)³ + 3(x − 1/x).

Best order to study it

Recall (a ± b)² and (x + a)(x + b); practise with coefficients and fractions.
Learn (a ± b)³ in both the long and short forms.
Learn (a ± b ± c)² and the sign rule for pairs.
Practise 'find the value' questions from sums and products.
Finish with x ± 1/x chains and quick arithmetic.

Quick revision checklist

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

Do you always include the middle term 2ab?
Do you square coefficients as well as variables?
Can you write (a − b)³ with the signs + − + −?
Can you expand a trinomial square with mixed signs?
Can you get a² + b², a − b and a³ + b³ from a + b and ab?
Can you move between x + 1/x and x − 1/x?
Can you work out 98², 103 × 104 and 99³ with identities?

Expansions: frequently asked questions

Is (a + b + c)³ in the Class IX syllabus?

No. The syllabus lists (a ± b)², (a ± b)³, (x ± a)(x ± b) and (a ± b ± c)². Stick to these.

What is the difference between an identity and an equation?

An identity is true for all values of the variable, such as (x + 1)² = x² + 2x + 1. An equation is true only for particular values, such as x + 1 = 3.

Should I use the long or short form of (a + b)³?

Use the long form a³ + 3a²b + 3ab² + b³ to expand, and the short form a³ + b³ + 3ab(a + b) to find values from a + b and ab.

Why does x + 1/x work so neatly?

Because x × 1/x = 1, so every ab term in the identities becomes 1 and drops out of the working.

Can I check an expansion quickly?

Yes. Put a simple value such as x = 1 into both sides. If the sides differ, there is an error.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028