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Expansions — ICSE Class 9 Maths Important Questions

13 ICSE Class 9 Maths practice questions on Expansions, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
Questions
5
Topics
3
Key concepts
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With answers
Quick answer

ICSE Expansions questions use the identities (a±b)2=a2±2ab+b2(a\pm b)^2=a^2\pm2ab+b^2(a± b)²=a²±2ab+b², (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), and (a±b)3=a3±3a2b+3ab2±b3(a\pm b)^3=a^3\pm3a^2b+3ab^2\pm b^3(a± b)³=a³±3a²b+3ab²± b³. Common tasks are expanding expressions, evaluating squares/cubes of numbers, and finding a2+1a2a^2+\dfrac{1}{a^2}a²+1/a² or a3+1a3a^3+\dfrac{1}{a^3}a³+1/a³ from a given value.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Expansions

In the ICSE Class 9 Maths chapter Expansions you apply standard algebraic identities for squares and cubes of binomials, the square of a trinomial, and product identities to expand expressions and to evaluate numerical powers and symmetric expressions such as x2+1x2x^2+\dfrac{1}{x^2}x²+1/x² efficiently.

Square of a binomialSquare of a trinomialCube of a binomialProduct identitiesEvaluating symmetric expressions

Key concepts & formulas

Squares

(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 (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).

Cubes

(a+b)3=a3+3a2b+3ab2+b3=a3+b3+3ab(a+b)(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)(a+b)³=a³+3a²b+3ab²+b³=a³+b³+3ab(a+b); (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).

Useful results

a2+1a2=(a+1a)2−2a^2+\dfrac{1}{a^2}=\left(a+\dfrac1a\right)^2-2a²+1/a²=(a+1/a)²-2 and a3+1a3=(a+1a)3−3(a+1a)a^3+\dfrac{1}{a^3}=\left(a+\dfrac1a\right)^3-3\left(a+\dfrac1a\right)a³+1/a³=(a+1/a)³-3(a+1/a).

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The expansion of (x+3)2(x+3)^2(x+3)² is:

  1. (a)

    x2+6x+9x^2+6x+9x²+6x+9

  2. (b)

    x2+9x^2+9x²+9

  3. (c)

    x2+3x+9x^2+3x+9x²+3x+9

  4. (d)

    x2+6x+6x^2+6x+6x²+6x+6

Show model answer

Answer: (a) x2+6x+9x^2+6x+9x²+6x+9.

(x+3)2=x2+2(x)(3)+32=x2+6x+9.(x+3)^2=x^2+2(x)(3)+3^2=x^2+6x+9.(x+3)²=x²+2(x)(3)+3²=x²+6x+9.

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Q2MCQEasy1 mark

The value of (a−b)3(a-b)^3(a-b)³ is:

  1. (a)

    a3−b3a^3-b^3a³-b³

  2. (b)

    a3−3a2b+3ab2−b3a^3-3a^2b+3ab^2-b^3a³-3a²b+3ab²-b³

  3. (c)

    a3+3a2b+3ab2+b3a^3+3a^2b+3ab^2+b^3a³+3a²b+3ab²+b³

  4. (d)

    a3−b3−3aba^3-b^3-3aba³-b³-3ab

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Answer: (b) a3−3a2b+3ab2−b3a^3-3a^2b+3ab^2-b^3a³-3a²b+3ab²-b³.

This is the standard expansion of the cube of a difference.

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Q3MCQModerate1 mark

If a+b=7a+b=7a+b=7 and ab=10ab=10ab=10, then a2+b2a^2+b^2a²+b² equals:

  1. (a)

    292929

  2. (b)

    494949

  3. (c)

    393939

  4. (d)

    595959

Show model answer

Answer: (a) 292929.

a2+b2=(a+b)2−2ab=49−20=29.a^2+b^2=(a+b)^2-2ab=49-20=29.a²+b²=(a+b)²-2ab=49-20=29.

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Q4MCQHOTS1 mark

If x+1x=3x+\dfrac1x=3x+1/x=3, then x3+1x3x^3+\dfrac{1}{x^3}x³+1/x³ equals:

  1. (a)

    272727

  2. (b)

    181818

  3. (c)

    999

  4. (d)

    212121

Show model answer

Answer: (b) 181818.

x3+1x3=(x+1x)3−3(x+1x)=27−9=18.x^3+\dfrac{1}{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)=27-9=18.x³+1/x³=(x+1/x)³-3(x+1/x)=27-9=18.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): (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).

Reason (R): The square of a sum of three terms contains the square of each term plus twice the product of each distinct pair.

  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: (a) Both are true and R correctly explains A. Expanding (a+b+c)2(a+b+c)^2(a+b+c)² produces each square once and each cross-product twice, matching the stated identity.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Evaluate (103)2(103)^2(103)² using a suitable identity.

Show model answer

Write 103=100+3103=100+3103=100+3 and use (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2(a+b)²=a²+2ab+b²:

(103)2=(100+3)2=1002+2(100)(3)+32=10000+600+9=10609.(103)^2=(100+3)^2=100^2+2(100)(3)+3^2=10000+600+9=10609.(103)²=(100+3)²=100²+2(100)(3)+3²=10000+600+9=10609.

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Q7Very ShortModerate2 marks

Expand (2x−12x)2\left(2x-\dfrac{1}{2x}\right)^2(2x-1/2x)².

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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=12xa=2x,\ b=\dfrac{1}{2x}a=2x, b=1/2x:

(2x−12x)2=(2x)2−2(2x)(12x)+(12x)2=4x2−2+14x2.\left(2x-\dfrac{1}{2x}\right)^2=(2x)^2-2(2x)\left(\dfrac{1}{2x}\right)+\left(\dfrac{1}{2x}\right)^2=4x^2-2+\dfrac{1}{4x^2}.(2x-1/2x)²=(2x)²-2(2x)(1/2x)+(1/2x)²=4x²-2+1/4x².

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

If a−1a=4a-\dfrac1a=4a-1/a=4, find a2+1a2a^2+\dfrac{1}{a^2}a²+1/a² and a4+1a4a^4+\dfrac{1}{a^4}a⁴+1/a⁴.

Show model answer

Square the given relation:

(a−1a)2=a2−2+1a2=16⇒a2+1a2=16+2=18.\left(a-\dfrac1a\right)^2=a^2-2+\dfrac{1}{a^2}=16\Rightarrow a^2+\dfrac{1}{a^2}=16+2=18.(a-1/a)²=a²-2+1/a²=16⇒ a²+1/a²=16+2=18.

Now square this result:

(a2+1a2)2=a4+2+1a4=182=324.\left(a^2+\dfrac{1}{a^2}\right)^2=a^4+2+\dfrac{1}{a^4}=18^2=324.(a²+1/a²)²=a⁴+2+1/a⁴=18²=324.

∴ a4+1a4=324−2=322.\therefore\ a^4+\dfrac{1}{a^4}=324-2=322.∴ a⁴+1/a⁴=324-2=322.

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Q9Short AnswerModerate3 marks

Expand (2x+3y−z)2(2x+3y-z)^2(2x+3y-z)².

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Use (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) with a=2x, b=3y, c=−za=2x,\ b=3y,\ c=-za=2x, b=3y, c=-z:

a2+b2+c2=4x2+9y2+z2a^2+b^2+c^2=4x^2+9y^2+z^2a²+b²+c²=4x²+9y²+z².

2(ab+bc+ca)=2[(2x)(3y)+(3y)(−z)+(−z)(2x)]=2[6xy−3yz−2zx]=12xy−6yz−4zx2(ab+bc+ca)=2\big[(2x)(3y)+(3y)(-z)+(-z)(2x)\big]=2[6xy-3yz-2zx]=12xy-6yz-4zx2(ab+bc+ca)=2[(2x)(3y)+(3y)(-z)+(-z)(2x)]=2[6xy-3yz-2zx]=12xy-6yz-4zx.

∴ (2x+3y−z)2=4x2+9y2+z2+12xy−6yz−4zx.\therefore\ (2x+3y-z)^2=4x^2+9y^2+z^2+12xy-6yz-4zx.∴ (2x+3y-z)²=4x²+9y²+z²+12xy-6yz-4zx.

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Q10Short AnswerHOTS3 marks

If x+y=5x+y=5x+y=5 and xy=6xy=6xy=6, find the value of x3+y3x^3+y^3x³+y³.

Show model answer

Use the identity x3+y3=(x+y)3−3xy(x+y)x^3+y^3=(x+y)^3-3xy(x+y)x³+y³=(x+y)³-3xy(x+y).

(x+y)3=53=125(x+y)^3=5^3=125(x+y)³=5³=125.

3xy(x+y)=3×6×5=903xy(x+y)=3\times6\times5=903xy(x+y)=3×6×5=90.

∴ x3+y3=125−90=35.\therefore\ x^3+y^3=125-90=35.∴ x³+y³=125-90=35.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

(a) If a+b+c=9a+b+c=9a+b+c=9 and ab+bc+ca=23ab+bc+ca=23ab+bc+ca=23, find a2+b2+c2a^2+b^2+c^2a²+b²+c².

(b) Using (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, expand (i) (x+4)(x−7)(x+4)(x-7)(x+4)(x-7) and (ii) (2y+3)(2y−5)(2y+3)(2y-5)(2y+3)(2y-5).

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(a) (a+b+c)2=a2+b2+c2+2(ab+bc+ca)⇒81=a2+b2+c2+46⇒a2+b2+c2=35(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\Rightarrow81=a^2+b^2+c^2+46\Rightarrow a^2+b^2+c^2=35(a+b+c)²=a²+b²+c²+2(ab+bc+ca)⇒81=a²+b²+c²+46⇒ a²+b²+c²=35.

(b)(i) a=4, b=−7a=4,\ b=-7a=4, b=-7: (x+4)(x−7)=x2+(4−7)x+(4)(−7)=x2−3x−28(x+4)(x-7)=x^2+(4-7)x+(4)(-7)=x^2-3x-28(x+4)(x-7)=x²+(4-7)x+(4)(-7)=x²-3x-28.

(ii) Take 2y2y2y as the variable, a=3, b=−5a=3,\ b=-5a=3, b=-5: (2y+3)(2y−5)=(2y)2+(3−5)(2y)+(3)(−5)=4y2−4y−15(2y+3)(2y-5)=(2y)^2+(3-5)(2y)+(3)(-5)=4y^2-4y-15(2y+3)(2y-5)=(2y)²+(3-5)(2y)+(3)(-5)=4y²-4y-15.

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Q12Long AnswerHOTS5 marks

If x+1x=4x+\dfrac1x=4x+1/x=4, evaluate (i) x2+1x2x^2+\dfrac{1}{x^2}x²+1/x², (ii) x3+1x3x^3+\dfrac{1}{x^3}x³+1/x³, and (iii) x4+1x4x^4+\dfrac{1}{x^4}x⁴+1/x⁴.

Show model answer

(i) Square x+1x=4x+\dfrac1x=4x+1/x=4:

(x+1x)2=x2+2+1x2=16⇒x2+1x2=16−2=14.\left(x+\dfrac1x\right)^2=x^2+2+\dfrac{1}{x^2}=16\Rightarrow x^2+\dfrac{1}{x^2}=16-2=14.(x+1/x)²=x²+2+1/x²=16⇒ x²+1/x²=16-2=14.

(ii) Cube x+1x=4x+\dfrac1x=4x+1/x=4:

x3+1x3=(x+1x)3−3(x+1x)=64−12=52.x^3+\dfrac{1}{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)=64-12=52.x³+1/x³=(x+1/x)³-3(x+1/x)=64-12=52.

(iii) Square the result of (i):

(x2+1x2)2=x4+2+1x4=142=196⇒x4+1x4=196−2=194.\left(x^2+\dfrac{1}{x^2}\right)^2=x^4+2+\dfrac{1}{x^4}=14^2=196\Rightarrow x^4+\dfrac{1}{x^4}=196-2=194.(x²+1/x²)²=x⁴+2+1/x⁴=14²=196⇒ x⁴+1/x⁴=196-2=194.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A student is asked to evaluate large numbers quickly using algebraic identities instead of long multiplication.

(i) Evaluate (98)2(98)^2(98)² using (a−b)2(a-b)^2(a-b)².

(ii) Evaluate (102)2(102)^2(102)² using (a+b)2(a+b)^2(a+b)².

(iii) Evaluate 105×95105\times95105×95 using (a+b)(a−b)(a+b)(a-b)(a+b)(a-b).

(iv) Evaluate (101)3(101)^3(101)³ using (a+b)3(a+b)^3(a+b)³.

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(i) (98)2=(100−2)2=10000−400+4=9604.(98)^2=(100-2)^2=10000-400+4=9604.(98)²=(100-2)²=10000-400+4=9604.

(ii) (102)2=(100+2)2=10000+400+4=10404.(102)^2=(100+2)^2=10000+400+4=10404.(102)²=(100+2)²=10000+400+4=10404.

(iii) 105×95=(100+5)(100−5)=1002−52=10000−25=9975.105\times95=(100+5)(100-5)=100^2-5^2=10000-25=9975.105×95=(100+5)(100-5)=100²-5²=10000-25=9975.

(iv) (101)3=(100+1)3=1003+3(1002)(1)+3(100)(12)+13=1000000+30000+300+1=1030301.(101)^3=(100+1)^3=100^3+3(100^2)(1)+3(100)(1^2)+1^3=1000000+30000+300+1=1030301.(101)³=(100+1)³=100³+3(100²)(1)+3(100)(1²)+1³=1000000+30000+300+1=1030301.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • What are the important expansions for ICSE Class 9?
    (a±b)2=a2±2ab+b2(a\pm b)^2=a^2\pm2ab+b^2(a± b)²=a²±2ab+b², (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), (a±b)3=a3±3a2b+3ab2±b3(a\pm b)^3=a^3\pm3a^2b+3ab^2\pm b^3(a± b)³=a³±3a²b+3ab²± b³, and the results a2+1a2=(a+1a)2−2a^2+\tfrac{1}{a^2}=\left(a+\tfrac1a\right)^2-2a²+1/a²=(a+1/a)²-2 and a3+1a3=(a+1a)3−3(a+1a)a^3+\tfrac{1}{a^3}=\left(a+\tfrac1a\right)^3-3\left(a+\tfrac1a\right)a³+1/a³=(a+1/a)³-3(a+1/a). This page has 13 questions that use them, including a case study on working out large numbers quickly without long multiplication.
  • If x + 1/x = 3, what is x³ + 1/x³?
    Use x3+1x3=(x+1x)3−3(x+1x)=27−9=18x^3+\dfrac{1}{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)=27-9=18x³+1/x³=(x+1/x)³-3(x+1/x)=27-9=18. This comes from cubing x+1xx+\dfrac1xx+1/x and rearranging, and it saves you from finding xxx itself. Questions in this style usually give one value of x+dfrac1xx+dfrac1xx+dfrac1x and ask for higher powers one after another.
  • How do you find x³ + y³ from x + y and xy?
    Use x3+y3=(x+y)3−3xy(x+y)x^3+y^3=(x+y)^3-3xy(x+y)x³+y³=(x+y)³-3xy(x+y). If x+y=5x+y=5x+y=5 and xy=6xy=6xy=6, then x3+y3=125−90=35x^3+y^3=125-90=35x³+y³=125-90=35. Similarly a2+b2=(a+b)2−2aba^2+b^2=(a+b)^2-2aba²+b²=(a+b)²-2ab, so a+b=7a+b=7a+b=7 and ab=10ab=10ab=10 give 292929. Both shortcuts save time in the exam and work even when the two numbers are not whole.
  • Which ICSE Class 9 chapter uses expansions next?
    Factorisation (Chapter 5) reverses these identities, so it follows Expansions (Chapter 4) directly. Both chapters have important-questions pages on Classmate AI, linked from the ICSE Class 9 Maths list.
  • How do I get faster at ICSE expansion questions?
    Memorise the identities in the form you will use them — for example (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 practise the a±1aa\pm\tfrac1aa± 1/a chain, where one value gives the square, cube and fourth-power sums in turn. The 5-mark question here runs that chain for x+1x=4x+\tfrac1x=4x+1/x=4.
  • Can I get ICSE Class 9 Expansions questions as a PDF?
    Yes. Enter your email in the free PDF box on this page to open a printable worksheet of all 13 Expansions questions with step-by-step answers, then save it as a PDF.
  • Can I take a timed test on ICSE Expansions?
    Yes. Sign up free, open Create Test, choose ICSE Class 9 Maths and Expansions, and set a timer and difficulty. Each test is marked instantly and recorded in your SWOT report, with two free sets every week.

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