Classmate AI
Learn · Solve · Practice
Chapter 4CBSE Class 9 Maths100% Free

Exploring Algebraic Identities — CBSE Class 9 Maths Important Questions

13 hand-picked CBSE Class 9 Maths important questions for Exploring Algebraic Identities, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

High-yield identity questions ask you to expand using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2(a+b)^2=a^2+2ab+b^2, (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2(a-b)^2=a^2-2ab+b^2, a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a^2-b^2=(a+b)(a-b), (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b)=x^2+(a+b)x+ab(x+a)(x+b)=x^2+(a+b)x+ab, (a+b+c)2(a+b+c)^2(a+b+c)^2, and (a±b)3(a\pm b)^3(a± b)^3. They also ask you to evaluate products like 103×97103\times97103×97 mentally, and to factorise. Choosing the right identity turns a long expansion into one step.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Exploring Algebraic Identities

Algebraic identities are equalities true for all values of the variables. This chapter collects the standard identities — the square of a binomial and trinomial, the difference of squares, the product (x+a)(x+b)(x+a)(x+b)(x+a)(x+b), and the cubes (a±b)3(a\pm b)^3(a± b)^3 — and uses them to expand expressions, evaluate numerical products quickly, and factorise. Mastering which identity fits saves time in almost every algebra problem.

Squares of binomials: $(a\pm b)^2$Difference of two squares: $a^2-b^2$Product $(x+a)(x+b)$Square of a trinomial: $(a+b+c)^2$Cubes of binomials and factorisation

Key concepts & formulas

Square identities

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)^2 = a^2 + 2ab + b^2, (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2(a-b)^2 = a^2 - 2ab + b^2, and a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b)a^2 - b^2 = (a+b)(a-b).

Trinomial square and product

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca; and (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b) = x^2 + (a+b)x + ab(x+a)(x+b) = x^2 + (a+b)x + ab.

Cube identities

(a+b)3=a3+b3+3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b) and (a−b)3=a3−b3−3ab(a−b)(a-b)^3 = a^3 - b^3 - 3ab(a-b)(a-b)^3 = a^3 - b^3 - 3ab(a-b); also a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a^3+b^3=(a+b)(a^2-ab+b^2).

Free download

Get all 13 Exploring Algebraic Identities questions as a PDF

The full question bank with model answers — perfect for offline revision and last-minute practice.

Free · No spam · Unsubscribe anytime

Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The expansion of (x+5)2(x + 5)^2(x + 5)^2 is:

  1. (a)

    x2+25x^2 + 25x^2 + 25

  2. (b)

    x2+10x+25x^2 + 10x + 25x^2 + 10x + 25

  3. (c)

    x2+5x+25x^2 + 5x + 25x^2 + 5x + 25

  4. (d)

    x2+10x+10x^2 + 10x + 10x^2 + 10x + 10

Show model answer

Answer: (b) x2+10x+25x^2 + 10x + 25x^2 + 10x + 25.

Using (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)^2 = a^2 + 2ab + b^2 with a=x, b=5a=x,\ b=5a=x, b=5: x2+2(x)(5)+52=x2+10x+25.x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25.x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25.

Still stuck? Ask the AI tutor to explain this step by step →
Q2MCQEasy1 mark

a2−b2a^2 - b^2a^2 - b^2 is equal to:

  1. (a)

    (a−b)2(a-b)^2(a-b)^2

  2. (b)

    (a+b)(a−b)(a+b)(a-b)(a+b)(a-b)

  3. (c)

    (a+b)2(a+b)^2(a+b)^2

  4. (d)

    a2+b2a^2 + b^2a^2 + b^2

Show model answer

Answer: (b) (a+b)(a−b)(a+b)(a-b)(a+b)(a-b).

This is the difference-of-squares identity: a2−b2=(a+b)(a−b).a^2 - b^2 = (a+b)(a-b).a^2 - b^2 = (a+b)(a-b).

Still stuck? Ask the AI tutor to explain this step by step →
Q3MCQModerate1 mark

Using an identity, the value of 103×97103 \times 97103 × 97 is:

  1. (a)

    999199919991

  2. (b)

    100001000010000

  3. (c)

    999999999999

  4. (d)

    100911009110091

Show model answer

Answer: (a) 999199919991.

Write 103×97=(100+3)(100−3)=1002−32=10000−9=9991103\times97 = (100+3)(100-3) = 100^2 - 3^2 = 10000 - 9 = 9991103×97 = (100+3)(100-3) = 100^2 - 3^2 = 10000 - 9 = 9991, using a2−b2=(a+b)(a−b).a^2-b^2=(a+b)(a-b).a^2-b^2=(a+b)(a-b).

Still stuck? Ask the AI tutor to explain this step by step →
Q4MCQHOTS1 mark

If x+1x=5x + \dfrac{1}{x} = 5x + 1/x = 5, then x2+1x2x^2 + \dfrac{1}{x^2}x^2 + 1/x^2 equals:

  1. (a)

    252525

  2. (b)

    232323

  3. (c)

    272727

  4. (d)

    101010

Show model answer

Answer: (b) 232323.

Square both sides: (x+1x)2=x2+2+1x2=25\left(x+\tfrac1x\right)^2 = x^2 + 2 + \tfrac{1}{x^2} = 25(x+1x)^2 = x^2 + 2 + 1x^2 = 25. Hence x2+1x2=25−2=23.x^2 + \tfrac{1}{x^2} = 25 - 2 = 23.x^2 + 1x^2 = 25 - 2 = 23.

Still stuck? Ask the AI tutor to explain this step by step →

Want every Exploring Algebraic Identities question solved live, at your pace?

Practise free with the AI tutor →

Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): (a+b)2=a2+b2(a+b)^2 = a^2 + b^2(a+b)^2 = a^2 + b^2 for all real a,ba, ba, b.

Reason (R): (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)^2 = a^2 + 2ab + b^2.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (d) A is false but R is true. The correct identity is (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)^2 = a^2 + 2ab + b^2 (R), which contains the extra middle term 2ab2ab2ab. So A, which omits 2ab2ab2ab, is false.

Still stuck? Ask the AI tutor to explain this step by step →

Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Expand (2x−3y)2(2x - 3y)^2(2x - 3y)^2 using a suitable identity.

Show model answer

Using (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2(a-b)^2 = a^2 - 2ab + b^2 with a=2x, b=3ya = 2x,\ b = 3ya = 2x, b = 3y:

(2x−3y)2=(2x)2−2(2x)(3y)+(3y)2=4x2−12xy+9y2.(2x-3y)^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = 4x^2 - 12xy + 9y^2.(2x-3y)^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = 4x^2 - 12xy + 9y^2.

Still stuck? Ask the AI tutor to explain this step by step →
Q7Very ShortModerate2 marks

Find the product (x+7)(x−4)(x + 7)(x - 4)(x + 7)(x - 4) using the identity for (x+a)(x+b)(x+a)(x+b)(x+a)(x+b).

Show model answer

Here a=7, b=−4a = 7,\ b = -4a = 7, b = -4, so a+b=3a+b = 3a+b = 3 and ab=−28ab = -28ab = -28.

Using (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b) = x^2 + (a+b)x + ab(x+a)(x+b) = x^2 + (a+b)x + ab:

(x+7)(x−4)=x2+3x−28.(x+7)(x-4) = x^2 + 3x - 28.(x+7)(x-4) = x^2 + 3x - 28.

Still stuck? Ask the AI tutor to explain this step by step →

Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Expand (2a+3b+c)2(2a + 3b + c)^2(2a + 3b + c)^2 using the identity for the square of a trinomial.

Show model answer

Use (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca with terms 2a, 3b, c2a,\ 3b,\ c2a, 3b, c.

Squares: (2a)2=4a2(2a)^2 = 4a^2(2a)^2 = 4a^2, (3b)2=9b2(3b)^2 = 9b^2(3b)^2 = 9b^2, (c)2=c2(c)^2 = c^2(c)^2 = c^2.

Cross terms: 2(2a)(3b)=12ab2(2a)(3b) = 12ab2(2a)(3b) = 12ab, 2(3b)(c)=6bc2(3b)(c) = 6bc2(3b)(c) = 6bc, 2(c)(2a)=4ca2(c)(2a) = 4ca2(c)(2a) = 4ca.

Therefore (2a+3b+c)2=4a2+9b2+c2+12ab+6bc+4ca.(2a+3b+c)^2 = 4a^2 + 9b^2 + c^2 + 12ab + 6bc + 4ca.(2a+3b+c)^2 = 4a^2 + 9b^2 + c^2 + 12ab + 6bc + 4ca.

Still stuck? Ask the AI tutor to explain this step by step →
Q9Short AnswerModerate3 marks

Factorise: x2+7x+12x^2 + 7x + 12x^2 + 7x + 12.

Show model answer

We need two numbers whose product is 121212 (the constant) and whose sum is 777 (the coefficient of xxx).

The numbers 333 and 444 work, since 3×4=123\times4 = 123×4 = 12 and 3+4=73+4 = 73+4 = 7.

Split the middle term: x2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4).x^2 + 3x + 4x + 12 = x(x+3) + 4(x+3) = (x+3)(x+4).x^2 + 3x + 4x + 12 = x(x+3) + 4(x+3) = (x+3)(x+4).

So x2+7x+12=(x+3)(x+4).x^2 + 7x + 12 = (x+3)(x+4).x^2 + 7x + 12 = (x+3)(x+4).

Still stuck? Ask the AI tutor to explain this step by step →
Q10Short AnswerHOTS3 marks

Without actually cubing, evaluate 98398^398^3 using the identity (a−b)3=a3−b3−3ab(a−b)(a-b)^3 = a^3 - b^3 - 3ab(a-b)(a-b)^3 = a^3 - b^3 - 3ab(a-b).

Show model answer

Write 98=100−298 = 100 - 298 = 100 - 2, so a=100, b=2a = 100,\ b = 2a = 100, b = 2.

a3=1000000a^3 = 1000000a^3 = 1000000, b3=8b^3 = 8b^3 = 8, and 3ab(a−b)=3(100)(2)(98)=600×98=588003ab(a-b) = 3(100)(2)(98) = 600\times 98 = 588003ab(a-b) = 3(100)(2)(98) = 600× 98 = 58800.

Then (100−2)3=a3−b3−3ab(a−b)=1000000−8−58800=941192.(100-2)^3 = a^3 - b^3 - 3ab(a-b) = 1000000 - 8 - 58800 = 941192.(100-2)^3 = a^3 - b^3 - 3ab(a-b) = 1000000 - 8 - 58800 = 941192.

So 983=941192.98^3 = 941192.98^3 = 941192.

Still stuck? Ask the AI tutor to explain this step by step →

Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Prove geometrically or algebraically that (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)^2 = a^2 + 2ab + b^2, and use it to expand (3x+4)2(3x + 4)^2(3x + 4)^2.

Show model answer

Algebraic proof: (a+b)2=(a+b)(a+b)=a(a+b)+b(a+b)=a2+ab+ba+b2=a2+2ab+b2.(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2.(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2.

Geometric view: A square of side a+ba+ba+b splits into four regions — a square a2a^2a^2, two rectangles each of area ababab, and a square b2b^2b^2 — so the total area is a2+2ab+b2a^2 + 2ab + b^2a^2 + 2ab + b^2.

CBSE Class 9 Maths — Exploring Algebraic Identities: Prove geometrically or algebraically that (a+b)^2 = a^2 + 2ab + b^2, and use it to expand (3x + 4)^2.

Application: With a=3x, b=4a = 3x,\ b = 4a = 3x, b = 4: (3x+4)2=(3x)2+2(3x)(4)+42=9x2+24x+16.(3x+4)^2 = (3x)^2 + 2(3x)(4) + 4^2 = 9x^2 + 24x + 16.(3x+4)^2 = (3x)^2 + 2(3x)(4) + 4^2 = 9x^2 + 24x + 16.

Still stuck? Ask the AI tutor to explain this step by step →
Q12Long AnswerHOTS5 marks

If a+b+c=9a + b + c = 9a + b + c = 9 and ab+bc+ca=23ab + bc + ca = 23ab + bc + ca = 23, find the value of a2+b2+c2a^2 + b^2 + c^2a^2 + b^2 + c^2. Then, if additionally abc=15abc = 15abc = 15, comment on how you would find a3+b3+c3a^3+b^3+c^3a^3+b^3+c^3.

Show model answer

Use the identity (a+b+c)2=a2+b2+c2+2(ab+bc+ca).(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca).(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca).

Substitute the given values:

92=(a2+b2+c2)+2(23).9^2 = (a^2+b^2+c^2) + 2(23).9^2 = (a^2+b^2+c^2) + 2(23).

81=(a2+b2+c2)+46.81 = (a^2+b^2+c^2) + 46.81 = (a^2+b^2+c^2) + 46.

Therefore a2+b2+c2=81−46=35.a^2 + b^2 + c^2 = 81 - 46 = 35.a^2 + b^2 + c^2 = 81 - 46 = 35.

Extending to cubes: Use the identity
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).

Here a2+b2+c2−(ab+bc+ca)=35−23=12a^2+b^2+c^2 - (ab+bc+ca) = 35 - 23 = 12a^2+b^2+c^2 - (ab+bc+ca) = 35 - 23 = 12, so the right side is 9×12=1089\times 12 = 1089× 12 = 108. Then
a3+b3+c3=108+3abc=108+3(15)=108+45=153.a^3+b^3+c^3 = 108 + 3abc = 108 + 3(15) = 108 + 45 = 153.a^3+b^3+c^3 = 108 + 3abc = 108 + 3(15) = 108 + 45 = 153.

Still stuck? Ask the AI tutor to explain this step by step →

Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A square garden has each side of length (x+6)(x + 6)(x + 6) metres. A square flower-bed of side xxx metres is left inside it, and the border (the region between the two squares) is paved.

(i) Write an expression for the total area of the garden.

(ii) Write an expression for the area of the flower-bed.

(iii) Using a suitable identity, find the paved border area in expanded form.

(iv) If x=10x = 10x = 10 m, find the numerical paved area.

Show model answer

(i) Garden area =(x+6)2= (x+6)^2= (x+6)^2 square metres.

(ii) Flower-bed area =x2= x^2= x^2 square metres.

(iii) Paved border area =(x+6)2−x2= (x+6)^2 - x^2= (x+6)^2 - x^2. Using (x+6)2=x2+12x+36(x+6)^2 = x^2 + 12x + 36(x+6)^2 = x^2 + 12x + 36:

(x+6)2−x2=(x2+12x+36)−x2=12x+36.(x+6)^2 - x^2 = (x^2 + 12x + 36) - x^2 = 12x + 36.(x+6)^2 - x^2 = (x^2 + 12x + 36) - x^2 = 12x + 36.

(Equivalently, by difference of squares, =(x+6+x)(x+6−x)=(2x+6)(6)=12x+36= (x+6+x)(x+6-x) = (2x+6)(6) = 12x+36= (x+6+x)(x+6-x) = (2x+6)(6) = 12x+36.)

(iv) At x=10x = 10x = 10: paved area =12(10)+36=120+36=156 m2.= 12(10) + 36 = 120 + 36 = 156\ \text{m}^2.= 12(10) + 36 = 120 + 36 = 156 m^2.

Still stuck? Ask the AI tutor to explain this step by step →

All CBSE Class 9 Maths Chapters

Orienting Yourself: The Use of CoordinatesIntroduction to Linear PolynomialsThe World of NumbersExploring Algebraic IdentitiesI'm Up and Down, and Round and RoundMeasuring Space: Perimeter and AreaThe Mathematics of Maybe: Introduction to ProbabilityPredicting What Comes Next: Exploring Sequences and Progressions

Related study guide

Polynomials Class 9 CBSE: All Formulas + Important Questions Solved
Identities, zeroes, factor theorem — the complete Class 9 polynomials guide with worked CBSE questions.

Frequently asked questions

  • Are these Exploring Algebraic Identities important questions free?
    Yes. All 13 CBSE Class 9 Maths important questions for Exploring Algebraic Identities are free, with full model answers and no login required.
  • Do these Exploring Algebraic Identities questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Exploring Algebraic Identities important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Exploring Algebraic Identities?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

Stuck on Exploring Algebraic Identities? Let the AI tutor help

Free to start · Step-by-step Socratic help · CBSE Class 9 Maths

Practise Exploring Algebraic Identities free →
💬 Share on WhatsApp
Practise Exploring Algebraic Identities free with the AI tutor →