MathsCBSE Class 8 · We Distribute, Yet Things Multiply

Algebraic Identities and Fast Multiplication for Class 8

Updated 17 October 2026 · Fact-checked

The standard identities are (a + b)^2 = a^2 + 2ab + b^2, (a - b)^2 = a^2 - 2ab + b^2 and (a + b)(a - b) = a^2 - b^2. They give fast ways to square and multiply, such as 104^2 = (100 + 4)^2, 97 x 103 = 100^2 - 3^2, and Sridharacharya's a^2 = (a + b)(a - b) + b^2. Exams ask for expansion, mental calculation, proofs of number patterns and area problems.

Notice the big square holds a², b² and two ab rectangles, so (a + b)² has a middle term 2ab and is not just a² + b².

Illustration drawn for QuestionBuddy.

Understand Algebraic Identities and Fast Multiplication

The distributive property gives three identities (Identity 1A, 1B and 1C in the chapter) that save a great deal of work.

Identity 1A: square of a sum. $(a + b)^{2} = (a + b) (a + b) = a^{2} + ab + ba + b^{2} = a^{2} + 2 ab + b^{2}$ . Geometrically, a square of side $a + b$ is made of a square of side $a$ , a square of side $b$ and two $a \times b$ rectangles. For example $6 5^{2} = (60 + 5)^{2} = 3600 + 2 \times 60 \times 5 + 25 = 4225$ . Note that $(a + b)^{2}$ is not $a^{2} + b^{2}$ ; the extra $2 ab$ is the two rectangles.

Identity 1B: square of a difference. $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ . It follows from 1A by writing $a - b = a + (- b)$ . Geometrically, remove two rectangles of $a \times b$ from the square of side $a$ , which removes the small $b \times b$ square twice, so add it back once. Hence $(a - b)^{2}$ and $(b - a)^{2}$ are equal (and both are never negative).

Identity 1C: difference of squares. $(a + b) (a - b) = a^{2} - ab + ba - b^{2} = a^{2} - b^{2}$ . For example, $98 \times 102 = (100 - 2) (100 + 2) = 10000 - 4 = 9996$ .

Two useful consequences.

$(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$ , so twice the sum of two squares is also a sum of two squares: $2 (5^{2} + 6^{2}) = 1 1^{2} + 1^{2}$ .
*Sridharacharya's method* (c. 750 CE in the chapter) rewrites 1C as $a^{2} = (a + b) (a - b) + b^{2}$ . Choose $b$ so that $a - b$ or $a + b$ is a round number: $3 1^{2} = 32 \times 30 + 1 = 961$ .

Fast multiplication. The distributive property gives quick one-line methods:

*Multiply by 11:* $d c ba \times 11 = d c ba \times 10 + d c ba$ . Write the units digit, then add each digit to its right-hand neighbour, carrying as needed, and finish with the leading digit.
*Multiply by 101, 1001, ...:* $N \times 101 = N \times 100 + N$ .
*Multiply by 99, 999, ...:* $N \times 99 = N \times 100 - N$ .

The chapter notes that similar methods appear in the works of Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE).

Areas. Many geometric areas are differences of squares or rectangles, for example a square of side $m + n$ with a square of side $n - m$ removed, which leaves $4 mn$ . Different but correct ways of finding an area give expressions that are equal after expansion.

Checking simplifications. Wrong simplifications such as $(a + b)^{2} = a^{2} + b^{2}$ are caught by substituting small numbers.

The distributive property itself and general expansion are on the previous topic page of this chapter.

Algebraic Identities and Fast Multiplication: formulas to remember

Identity 1A: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$; Square of a sum.
Identity 1B: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$; Square of a difference.
Identity 1C: $(a + b) (a - b) = a^{2} - b^{2}$; Product of a sum and a difference.
Sum of two squares, doubled: $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$; From adding 1A and 1B.
Sridharacharya form: $a^{2} = (a + b) (a - b) + b^{2}$; Choose b to make a + b or a - b a round number.
Four times a product: $(a + b)^{2} - (a - b)^{2} = 4 ab$; Subtracting 1B from 1A.

How to solve Algebraic Identities and Fast Multiplication questions

Pick the identity from the shape of the question.

1If you see a bracket squared, use 1A or 1B; match a and b with the two terms and keep their signs.
2If you see (term + term)(term - term), use 1C and square each term.
3For mental calculation, split the number around a round number (100, 50, 1000) and decide between 1A, 1B and 1C.
4For number patterns, replace the numbers by letter-numbers and expand both sides.
5For areas, write the area as outer minus inner, or as a sum of pieces, then simplify using identities.
6Check the result with a small substitution such as a = 1, b = 2.

Algebraic Identities and Fast Multiplication in real life

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 floor of a hall is a square of 65 tiles by 65 tiles and you want to count the tiles without a calculator.

Find the number of tiles.

Try it first, then show the solutionSolution

$65 = 60 + 5$ , so $6 5^{2} = 6 0^{2} + 2 \times 60 \times 5 + 5^{2}$ .
$= 3600 + 600 + 25 = 4225$ .

Answer: 4225 tiles

Memory hook: Picture the 65 by 65 square as a 60 by 60 square, two strips and a small corner square.

Difference of squares in area

Imagine a square garden of side 45 m with a square pond of side 35 m in the middle, leaving a path of equal width all round.

Find the area of the path and its width.

Try it first, then show the solutionSolution

Area of path $= 4 5^{2} - 3 5^{2} = (45 + 35) (45 - 35) = 80 \times 10 = 800$ m $^{2}$ .
Width $= (45 - 35) \div 2 = 5$ m.

Answer: 800 square metres; 5 m wide

Memory hook: Outer square minus inner square is a difference of squares.

Tricks to remember Algebraic Identities and Fast Multiplication

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

Square the first, twice the product, square the last
For (a + b) squared: a squared, plus 2ab, plus b squared. Change the middle sign for a difference.
Use it when: Expanding squares quickly.
Middle terms cancel
In (a + b)(a - b) the plus ab and minus ab cancel, leaving only the squares.
Use it when: Spotting the use of 1C.
Round-number trick
Split the number as hundred plus or minus something small, then pick the identity.
Use it when: Mental squares such as 104 squared and 97 times 103.
Square minus b squared
Sridharacharya: a squared equals (a plus b)(a minus b) plus b squared.
Use it when: Squaring numbers near a round number.

Algebraic Identities and Fast Multiplication: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Expanding a square · 1 mark · MCQBoard level

The expansion of $(2 a - 3)^{2}$ is: (a) $4 a^{2} - 9$ (b) $4 a^{2} - 6 a + 9$ (c) $4 a^{2} - 12 a + 9$ (d) $2 a^{2} - 12 a + 9$

Try it first, then show the solutionSolution

$(2 a - 3)^{2} = (2 a)^{2} - 2 (2 a) (3) + 3^{2} = 4 a^{2} - 12 a + 9$ .

Answer: (c) 4a^2 - 12a + 9

Example 2: Square by Identity 1A · 2 marks · Short answerBoard level

Find 104 squared using an identity.

Try it first, then show the solutionSolution

$104 = 100 + 4$ . (1 mark)
$10 4^{2} = 10 0^{2} + 2 \times 100 \times 4 + 4^{2} = 10000 + 800 + 16 = 10816$ . (1 mark)

Answer: 10816

Example 3: Product by Identity 1C · 2 marks · Short answerBoard level

Find 97 x 103 using an identity.

Try it first, then show the solutionSolution

$97 = 100 - 3$ and $103 = 100 + 3$ . (1 mark)
$97 \times 103 = 10 0^{2} - 3^{2} = 10000 - 9 = 9991$ . (1 mark)

Answer: 9991

Example 4: Multiplying by 11 in one line · 3 marks · Short answerHarder board level

Find 3279 x 11 by the one-line method and explain the working with the distributive property.

Try it first, then show the solutionSolution

$3279 \times 11 = 3279 \times 10 + 3279 = 32790 + 3279$ . (1 mark)
Units digit 9. Tens: $7 + 9 = 16$ , write 6 carry 1. Hundreds: $2 + 7 + 1 = 10$ , write 0 carry 1. Thousands: $3 + 2 + 1 = 6$ . Leading digit 3. (1 mark)
Result: 3, 6, 0, 6, 9 $= 36069$ . (1 mark)

Answer: 36069

Example 5: Sridharacharya squaring · 4 marks · Long answerHarder board level

Prove a^2 = (a + b)(a - b) + b^2. Use it to find 72 squared and 145 squared.

Try it first, then show the solutionSolution

$(a + b) (a - b) = a^{2} - b^{2}$ (Identity 1C), so $(a + b) (a - b) + b^{2} = a^{2}$ . (1 mark)
For 72 take $b = 2$ : $7 2^{2} = 74 \times 70 + 4 = 5180 + 4$ . (1 mark)
So $7 2^{2} = 5184$ . (1 mark)
For 145 take $b = 5$ : $14 5^{2} = 150 \times 140 + 25 = 21000 + 25 = 21025$ . (1 mark)

Answer: 72 squared = 5184; 145 squared = 21025

Where students lose marks in Algebraic Identities and Fast Multiplication

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

Mistake: Writing (a + b) squared as a squared plus b squared
The middle term 2ab is missing.
Wrong working
" $(x + 5)^{2} = x^{2} + 25$ "
Right working
$(x + 5)^{2} = x^{2} + 10 x + 25$ .
Do this: Test with x = 1: left is 36, wrong right is 26.
Usually costs: 2 marks.
Mistake: Wrong sign in the middle term of (a - b) squared
The middle term is negative, the last term positive.
Wrong working
" $(a - b)^{2} = a^{2} - 2 ab - b^{2}$ "
Right working
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
Do this: Remember that (-b) squared is positive.
Usually costs: 1 mark.
Mistake: Splitting numbers badly for mental calculation
Choose the split that gives an easy product, otherwise there is no time saved and errors creep in.
Wrong working
" $97 \times 103 = (90 + 7) (100 + 3)$ "
Right working
$97 \times 103 = (100 - 3) (100 + 3) = 9991$ .
Do this: Look for numbers equally spaced from a round number.
Usually costs: Method marks.
Mistake: Forgetting to add b squared in the Sridharacharya method
a squared equals the product plus b squared.
Wrong working
" $7 2^{2} = 74 \times 70 = 5180$ "
Right working
$7 2^{2} = 74 \times 70 + 2^{2} = 5184$ .
Do this: Write the identity before substituting.
Usually costs: The answer.

Exam tips

Name the identity you use (1A, 1B or 1C) and write the general form before substituting.
For mental products, show the split (for example 97 = 100 - 3) on one line, then the identity.
Check every expansion by putting a = 1, b = 1 or 2 into both sides.
For area questions, write the area as outer minus inner and then use 1C if both are squares.
Show the verification of a pattern with at least two examples after the algebraic proof.

Practice questions on Algebraic Identities and Fast Multiplication

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 · Assertion-reasonBoard level
Assertion (A): (a - b)^2 = (b - a)^2 for all numbers a and b.
Reason (R): (a - b)^2 = a^2 - 2ab + b^2 and (b - a)^2 = b^2 - 2ab + a^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.
Hint
Expand both sides.
Answer
(a). Both expansions are $a^{2} - 2 ab + b^{2}$ , so they are equal.
Q2 · 3 marks · Short answerHarder board level
Simplify $(2 a + 3 b)^{2} - (2 a - 3 b)^{2}$ and use the result to evaluate it for $a = 5$ , $b = 4$ .
Hint
Use (a + b)^2 - (a - b)^2 = 4ab, with the terms 2a and 3b.
Answer
$(2 a + 3 b)^{2} - (2 a - 3 b)^{2} = 4 (2 a) (3 b) = 24 ab$ . For $a = 5, b = 4$ : $24 \times 20 = 480$ . Check: $(10 + 12)^{2} - (10 - 12)^{2} = 484 - 4 = 480$ .
Q3 · 3 marks · Short answerHarder board level
Twice the sum of two squares is again a sum of two squares. Prove this with algebra and write 2(7^2 + 9^2) as a sum of two squares.
Hint
Add (a + b)^2 and (a - b)^2.
Answer
$(a + b)^{2} + (a - b)^{2} = (a^{2} + 2 ab + b^{2}) + (a^{2} - 2 ab + b^{2}) = 2 a^{2} + 2 b^{2} = 2 (a^{2} + b^{2})$ . For $a = 7, b = 9$ : $2 (49 + 81) = 260 = (7 + 9)^{2} + (9 - 7)^{2} = 1 6^{2} + 2^{2} = 256 + 4$ .
Q4 · 4 marks · Long answerHOTS / competency
Prove that the product of four consecutive integers plus 1 is always a perfect square. Verify with 1, 2, 3, 4 and with 2, 3, 4, 5.
Hint
Pair the outer two and the inner two numbers, and let t = n(n + 3).
Answer
$n (n + 1) (n + 2) (n + 3) = [n (n + 3)] [(n + 1) (n + 2)] = (n^{2} + 3 n) (n^{2} + 3 n + 2)$ . Let $t = n^{2} + 3 n$ . Then the product plus 1 is $t (t + 2) + 1 = t^{2} + 2 t + 1 = (t + 1)^{2} = (n^{2} + 3 n + 1)^{2}$ . Check: $n = 1$ : $24 + 1 = 25 = 5^{2}$ and $1 + 3 + 1 = 5$ . $n = 2$ : $120 + 1 = 121 = 1 1^{2}$ and $4 + 6 + 1 = 11$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a square garden has side 45 m and a square pond of side 35 m at its centre, with a path of uniform width w all round the pond.
(i) Find the area of the path using an identity. (1 mark)
(ii) Write the area of the path in terms of the garden side s and the width w, and show that it equals 4w(s - w). (2 marks)
(iii) If the pond were smaller, with side 31 m in the same garden, what would the path area be? (1 mark)
Hint
Pond side is s - 2w.
Answer
(i) $4 5^{2} - 3 5^{2} = (45 + 35) (45 - 35) = 80 \times 10 = 800$ m $^{2}$ . (ii) Pond side $= s - 2 w$ , so area of path $= s^{2} - (s - 2 w)^{2} = [s + (s - 2 w)] [s - (s - 2 w)] = (2 s - 2 w) (2 w) = 4 w (s - w)$ . With $s = 45$ , $w = 5$ : $4 \times 5 \times 40 = 800$ . (iii) $4 5^{2} - 3 1^{2} = (45 + 31) (45 - 31) = 76 \times 14 = 1064$ m $^{2}$ (width 7 m: $4 \times 7 \times 38 = 1064$ ).

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 (a + b) squared?Show answer
a squared + 2ab + b squared.
2.What is (a - b) squared?Show answer
a squared - 2ab + b squared.
3.What is (a + b)(a - b)?Show answer
a squared - b squared.
4.What is 2(a squared + b squared) in terms of (a + b) and (a - b)?Show answer
(a + b) squared + (a - b) squared.
5.How do you multiply a number by 11 quickly?Show answer
Number x 10 + number: add each digit to its neighbour from the right and carry.
6.Which ancient mathematician is linked with squaring using a squared = (a + b)(a - b) + b squared?Show answer
Sridharacharya (c. 750 CE) in the chapter.

Algebraic Identities and Fast Multiplication: frequently asked questions

Can I use these identities for negative numbers?

Yes. The identities hold for all integers, as the chapter checks.

Is (a + b) squared always greater than a squared plus b squared?

Not always. It exceeds it by 2ab, so it is greater only when ab is positive; for ab negative it is smaller, and for ab zero they are equal.

Do I have to learn the one-line method for 11?

It is a shortcut from the distributive property; understanding why it works matters more than memorising it.

How do I multiply 9734 by 99?

Use N x 99 = N x 100 - N: 973400 - 9734 = 963666.