CBSE • Class 9 • Mathematics • Chapter 2 (Polynomials) • Exercise 2.5

Exercise 2.5: Polynomials — NCERT Solutions

Algebraic identities — expanding and factorising cubic-type expressions.

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What this exercise covers

(a ± b)³a³ ± b³Special identities

Step-by-step solutions — Exercise 2.5

4 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 2.5 Q1 • 3 marks

Expand (2x+3y)3(2x + 3y)^3 using a suitable algebraic identity.
Hint (Socratic — try this first)▾
Which identity gives the expansion of (a+b)3(a + b)^3?
Step-by-step solution▾

Use the identity (a+b)3=a3+b3+3ab(a+b)(a + b)^3 = a^3 + b^3 + 3ab(a + b) with a=2xa = 2x and b=3yb = 3y.

a3=(2x)3=8x3,b3=(3y)3=27y3.a^3 = (2x)^3 = 8x^3, \qquad b^3 = (3y)^3 = 27y^3. 3ab(a+b)=3(2x)(3y)(2x+3y)=18xy(2x+3y)=36x2y+54xy2.3ab(a+b) = 3(2x)(3y)(2x + 3y) = 18xy(2x + 3y) = 36x^2 y + 54 x y^2.

Therefore: (2x+3y)3=8x3+27y3+36x2y+54xy2.(2x + 3y)^3 = 8x^3 + 27y^3 + 36x^2 y + 54 x y^2.

Common mistake:
Forgetting to cube the coefficients — e.g. writing (2x)3=2x3(2x)^3 = 2x^3 instead of 8x38x^3.
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Exercise 2.5 Q2 • 4 marks

Factorise x3−8y3+27z3+18xyzx^3 - 8y^3 + 27z^3 + 18xyz using a suitable identity.
Hint (Socratic — try this first)▾
Do the given terms fit the identity a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc for some a,b,ca, b, c?
Step-by-step solution▾

Rewrite the expression to match a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc.

Let a=xa = x, b=−2yb = -2y, c=3zc = 3z. Then: a3=x3,b3=(−2y)3=−8y3,c3=(3z)3=27z3.a^3 = x^3,\quad b^3 = (-2y)^3 = -8y^3,\quad c^3 = (3z)^3 = 27z^3. −3abc=−3(x)(−2y)(3z)=18xyz.-3abc = -3(x)(-2y)(3z) = 18xyz.

So the expression is exactly a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc.

Using 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):

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

Common mistake:
Sign errors when squaring/multiplying b=−2yb = -2y; for example writing the −ab-ab term as −2xy-2xy instead of +2xy+2xy.
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Exercise 2.5 Q3 • 3 marks

Without actually calculating the cubes, find the value of (−12)3+(7)3+(5)3(-12)^3 + (7)^3 + (5)^3.
Hint (Socratic — try this first)▾
What is special about the sum −12+7+5-12 + 7 + 5, and which identity does that trigger?
Step-by-step solution▾

Let a=−12a = -12, b=7b = 7, c=5c = 5. Notice: a+b+c=−12+7+5=0.a + b + c = -12 + 7 + 5 = 0.

When a+b+c=0a + b + c = 0, the identity a3+b3+c3−3abc=(a+b+c)(…)a^3 + b^3 + c^3 - 3abc = (a+b+c)(\ldots) gives a3+b3+c3=3abc.a^3 + b^3 + c^3 = 3abc.

Therefore: (−12)3+73+53=3(−12)(7)(5)=3×(−420)=−1260.(-12)^3 + 7^3 + 5^3 = 3(-12)(7)(5) = 3 \times (-420) = -1260.

Common mistake:
Ignoring that a+b+c=0a+b+c=0 and trying to cube each number directly, which is error-prone and misses the intended shortcut.
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Exercise 2.5 Q4 • 4 marks

Factorise the cubic polynomial x3−6x2+11x−6x^3 - 6x^2 + 11x - 6 completely.
Hint (Socratic — try this first)▾
Can you first spot one integer zero among the factors of the constant term, then divide?
Step-by-step solution▾

Let p(x)=x3−6x2+11x−6p(x) = x^3 - 6x^2 + 11x - 6. Try small factors of 66.

p(1)=1−6+11−6=0p(1) = 1 - 6 + 11 - 6 = 0, so (x−1)(x - 1) is a factor.

Divide p(x)p(x) by (x−1)(x - 1) to get the quotient: x3−6x2+11x−6=(x−1)(x2−5x+6).x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6).

Now factorise the quadratic by splitting the middle term: x2−5x+6=x2−2x−3x+6=(x−2)(x−3).x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = (x - 2)(x - 3).

Therefore: x3−6x2+11x−6=(x−1)(x−2)(x−3).x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3).

Common mistake:
Stopping after finding one factor (x−1)(x-1) and not factorising the resulting quadratic into two more linear factors.
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How to approach Exercise 2.5

  1. Re-read the chapter summary first. Open Polynomials and refresh the key concepts: Polynomial, Degree, Remainder theorem, Factor theorem.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Polynomials

  1. Exercise 2.1 — Polynomials in one variable — degree, coefficients and zeros.
  2. Exercise 2.2 — Zeros of linear and quadratic polynomials — finding roots algebraically.
  3. Exercise 2.3 — The Remainder Theorem — evaluating remainders without long division.
  4. Exercise 2.4 — The Factor Theorem — detecting factors (x − a) from functional values.
  5. Exercise 2.5 — Algebraic identities — expanding and factorising cubic-type expressions.

← Exercise 2.4

The Factor Theorem — detecting factors (x − a) from functional values.

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