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Exploring Algebraic Identities — CBSE Class 9 Maths Important Questions

13 CBSE Class 9 Maths practice questions on Exploring Algebraic Identities, each with a full model answer, covering 7 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
7
Topics
3
Key concepts
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Quick answer

Identity questions ask you to expand using (a±b)2(a\pm b)^2(a± b)², a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a²-b²=(a+b)(a-b), (x+a)(x+b)(x+a)(x+b)(x+a)(x+b), (a+b+c)2(a+b+c)^2(a+b+c)² and (a±b)3(a\pm b)^3(a± b)³, to see identities as areas of squares and rectangles, to factorise expressions with identities or algebra tiles, and to simplify rational expressions by cancelling common factors. They also ask you to evaluate products like 103×97103\times97103×97 mentally.

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 factorisationGeometric models and algebra tilesSimplifying rational expressions

Key concepts & formulas

Square identities

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2(a+b)² = a² + 2ab + b², (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2(a-b)² = a² - 2ab + b², and a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b)a² - b² = (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 = a²+b²+c²+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) = x² + (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 = a³ + b³ + 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 = a³ - b³ - 3ab(a-b); also a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²).

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Important questions with answers

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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 + 25x² + 25

  2. (b)

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

  3. (c)

    x2+5x+25x^2 + 5x + 25x² + 5x + 25

  4. (d)

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

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

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

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Q2MCQEasy1 mark

The factors of 9x2−169x^2-169x²-16 are:

  1. (a)

    (3x−4)2(3x-4)^2(3x-4)²

  2. (b)

    (3x+4)(3x−4)(3x+4)(3x-4)(3x+4)(3x-4)

  3. (c)

    (9x+4)(x−4)(9x+4)(x-4)(9x+4)(x-4)

  4. (d)

    (3x+16)(3x−1)(3x+16)(3x-1)(3x+16)(3x-1)

Show model answer

Answer: (b) (3x+4)(3x−4)(3x+4)(3x-4)(3x+4)(3x-4). 9x2−16=(3x)2−429x^2-16=(3x)^2-4^29x²-16=(3x)²-4², a difference of two squares, so it equals (3x+4)(3x−4)(3x+4)(3x-4)(3x+4)(3x-4).

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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) = 100² - 3² = 10000 - 9 = 9991, using a2−b2=(a+b)(a−b).a^2-b^2=(a+b)(a-b).a²-b²=(a+b)(a-b).

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Q4MCQHOTS1 mark

If x+1x=5x + \dfrac{1}{x} = 5x + 1/x = 5, then x2+1x2x^2 + \dfrac{1}{x^2}x² + 1/x² 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+1/x)² = x² + 2 + 1/x² = 25. Hence x2+1x2=25−2=23.x^2 + \tfrac{1}{x^2} = 25 - 2 = 23.x² + 1/x² = 25 - 2 = 23.

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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 = a² + b² 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 = a² + 2ab + b².

  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 = a² + 2ab + b² (R), which contains the extra middle term 2ab2ab2ab. So A, which omits 2ab2ab2ab, is false.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

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

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Using (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2(a-b)² = a² - 2ab + b² 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 = 4x² - 12xy + 9y².

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Q7Very ShortModerate2 marks

Simplify x2−9x2+5x+6\dfrac{x^2-9}{x^2+5x+6}(x²-9)/(x²+5x+6).

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Factorise the numerator and the denominator:
x2−9x2+5x+6=(x+3)(x−3)(x+2)(x+3)=x−3x+2,x≠−2, −3.\frac{x^2-9}{x^2+5x+6}=\frac{(x+3)(x-3)}{(x+2)(x+3)}=\frac{x-3}{x+2},\qquad x\neq-2,\ -3.(x²-9)/(x²+5x+6)=(x+3)(x-3)/(x+2)(x+3)=(x-3)/(x+2), x≠-2, -3.
The common factor (x+3)(x+3)(x+3) cancels; the values that make the denominator zero are excluded.

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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.

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

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

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 = 4a² + 9b² + c² + 12ab + 6bc + 4ca.

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Q9Short AnswerModerate3 marks

Factorise x2+7x+12x^2+7x+12x²+7x+12 by arranging algebra tiles (one x2x^2x² tile, seven xxx tiles and twelve unit tiles) into a rectangle. Check your answer by splitting the middle term.

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The tiles fit together into one rectangle:

CBSE Class 9 Maths — Exploring Algebraic Identities: Factorise x^2+7x+12 by arranging algebra tiles (one x^2 tile, seven x tiles and twelve unit tiles) into a rectangle. Check your

The sides of the rectangle are (x+4)(x+4)(x+4) and (x+3)(x+3)(x+3), so
x2+7x+12=(x+4)(x+3).x^2+7x+12=(x+4)(x+3).x²+7x+12=(x+4)(x+3).
Check: two numbers with product 121212 and sum 777 are 333 and 444: 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)x²+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4).

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Q10Short AnswerHOTS3 marks

Without actually cubing, evaluate 98398^398³ using the identity (a−b)3=a3−b3−3ab(a−b)(a-b)^3 = a^3 - b^3 - 3ab(a-b)(a-b)³ = a³ - b³ - 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 = 1000000a³ = 1000000, b3=8b^3 = 8b³ = 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 = a³ - b³ - 3ab(a-b) = 1000000 - 8 - 58800 = 941192.

So 983=941192.98^3 = 941192.98³ = 941192.

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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 = a² + 2ab + b², and use it to expand (3x+4)2(3x + 4)^2(3x + 4)².

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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) = a² + ab + ba + b² = a² + 2ab + b².

Geometric view: A square of side a+ba+ba+b splits into four regions — a square a2a^2a², two rectangles each of area ababab, and a square b2b^2b² — so the total area is a2+2ab+b2a^2 + 2ab + b^2a² + 2ab + b².

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) + 4² = 9x² + 24x + 16.

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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^2a²+b²+c². If also abc=15abc=15abc=15, find a3+b3+c3a^3+b^3+c^3a³+b³+c³.

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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 = a²+b²+c² + 2(ab+bc+ca).

Substitute the given values:

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

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

Therefore a2+b2+c2=81−46=35.a^2 + b^2 + c^2 = 81 - 46 = 35.a² + b² + c² = 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).a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² - ab - bc - ca).

Here a2+b2+c2−(ab+bc+ca)=35−23=12a^2+b^2+c^2 - (ab+bc+ca) = 35 - 23 = 12a²+b²+c² - (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.a³+b³+c³ = 108 + 3abc = 108 + 3(15) = 108 + 45 = 153.

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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.

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

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

(iii) Paved border area =(x+6)2−x2= (x+6)^2 - x^2= (x+6)² - x². Using (x+6)2=x2+12x+36(x+6)^2 = x^2 + 12x + 36(x+6)² = x² + 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 - x² = (x² + 12x + 36) - x² = 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 m².

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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

  • What extra questions come from Exploring Algebraic Identities Class 9?
    Expanding with (a±b)2(a\pm b)^2(a± b)², (a+b+c)2(a+b+c)^2(a+b+c)² and (a±b)3(a\pm b)^3(a± b)³, factorising with a2−b2a^2-b^2a²-b² or algebra tiles, simplifying rational expressions, and quick products like 103×97103\times97103×97. This page has 13 such questions with answers, including a square-garden case study.
  • How do you calculate 103 × 97 using an identity?
    Write it as (100+3)(100−3)(100+3)(100-3)(100+3)(100-3) and use (a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2(a+b)(a-b)=a²-b²: 1002−32=10000−9=9991100^2-3^2=10000-9=9991100²-3²=10000-9=9991. The same idea turns many products of numbers equally spaced around a round number into one subtraction.
  • If x + 1/x = 5, what is x² + 1/x²?
    Square both sides: x2+2+1x2=25x^2+2+\dfrac{1}{x^2}=25x²+2+1/x²=25, so x2+1x2=23x^2+\dfrac{1}{x^2}=23x²+1/x²=23. The middle term is 222 because 2⋅x⋅1x=22\cdot x\cdot\dfrac1x=22· x· 1/x=2; forgetting it is the most common error in this type of MCQ.
  • How do I find 98³ without multiplying it out?
    Use (a−b)3=a3−b3−3ab(a−b)(a-b)^3=a^3-b^3-3ab(a-b)(a-b)³=a³-b³-3ab(a-b) with a=100a=100a=100, b=2b=2b=2: 1000000−8−3×100×2×98=1000000−8−58800=9411921000000-8-3\times100\times2\times98=1000000-8-58800=9411921000000-8-3×100×2×98=1000000-8-58800=941192. Choosing aaa as a power of ten keeps every step mental, and the same method works for numbers just above a round value, such as 102102102 cubed with (a+b)3(a+b)^3(a+b)³.
  • Which Class 9 Ganita Manjari chapter covers algebraic identities?
    Exploring Algebraic Identities is Chapter 4 of the new CBSE Class 9 Maths book, after The World of Numbers (Chapter 3). Introduction to Linear Polynomials (Chapter 2) also has an important-questions page here if you want to revise polynomial basics first.
  • Why do I keep getting (a + b)² wrong?
    Usually by writing a2+b2a^2+b^2a²+b² and dropping 2ab2ab2ab — this page's assertion–reason question tests exactly that. Picturing a square of side a+ba+ba+b split into a2a^2a², b2b^2b² and two ababab rectangles fixes it; the 5-mark question asks you to prove it geometrically.
  • Is there a PDF of algebraic identities questions for Class 9?
    Yes. Enter your email in the free PDF box on this page and a printable sheet of the 13 identity questions with full solutions opens; choose Save as PDF to keep it.
  • Can I practise algebraic identities with an AI tutor?
    Yes. Press "Practise free" and Classmate AI's tutor gives you expressions to expand or factorise, checks which identity you chose, and hints at the next step when you are stuck. The Free plan includes three tutoring sessions a week.

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