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CBSE Class 9 Mathematics Question 6 of 16

Polynomials — Question 6

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Question

Question 6

Write the following cubes in expanded form:

(i) (2x + 1)3

(ii) (2a - 3b)3

(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3

(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3

Answer

(i) (2x + 1)3

We know that:

(a + b)3= (a)3 + (b)3 + 3a2b + 3ab2

Putting a = 2x, b = 1

= (2x)3 + (1)3 + 3(2x)2(1) + 3(2x)(1)2

= 8x3 + 12x2 + 6x + 1

Hence, (2x + 1)3 = 8x3 + 12x2 + 6x + 1

(ii) (2a - 3b)3

We know that:

(a - b)3= (a)3 - (b)3 - 3a2b + 3ab2

Putting a = 2a, b = 3b

= (2a)3 - (3b)3 - 3(2a)2(3b) + 3(2a)(3b)2

= 8a3 - 27b3 - 3(4a2)(3b) + 3(2a)(9b2)

= 8a3 - 27b3 - 36a2b + 54ab2

Hence, (2a - 3b)3 = 8a3 - 27b3 - 36a2b + 54ab2

(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3

We know that:

(a + b)3 = (a)3 + (b)3 + 3ab(a + b)

Putting a = 32x\dfrac{3}{2}x, b = 1

=(32)3+(1)3+3(3x2)(1)(3x2+1)=27x38+1+(9x2)(3x2+1)=27x38+1+(9x2)(3x2)+9x2(1)=27x38+27x24+9x2+1= \Big(\dfrac{3}{2}\Big)^3 + (1)^3 + 3 \Big(\dfrac{3x}{2}\Big)(1) \Big(\dfrac{3x}{2} + 1\Big) \\[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2} + 1\Big) \\[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2}\Big) + \dfrac{9x}{2}(1)\\[1em] = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1

Hence, [3x2+1]3=27x38+27x24+9x2+1\Big[\dfrac{3x}{2} + 1\Big]^3 = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1

(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3

We know that:

(a - b)3 = (a)3 - (b)3 - 3ab(a - b)

Putting a = x, b = −23y-\dfrac{2}{3}y

=(x)3−[−23y]3−3(x)[−23y][x−23y]=x3−827y3−2xy[x−23y]=x3−827y3−2x2y+43xy2= (x)^3 - \Big[-\dfrac{2}{3}y\Big]^3 - 3(x)\Big[-\dfrac{2}{3}y\Big] \Big[x - \dfrac{2}{3}y\Big]\\[1em] = x^3 - \dfrac{8}{27}y^3 - 2xy\Big[x - \dfrac{2}{3}y\Big]\\[1em] = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2

Hence, [x−23y]3=x3−827y3−2x2y+43xy2\Big[x - \dfrac{2}{3}y\Big]^3 = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2

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Polynomials — Interactive Study Guide

Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.

Polynomial Classification

Not a polynomial: √x (fractional power), 1/x = x−1 (negative power), x + 1/x.

Polynomial: 5 (constant), 3x + 2 (linear), x² − 1 (quadratic), 2x³ + x − 1 (cubic).

Remainder and Factor Theorems — Quick Guide

Remainder Theorem: When p(x) is divided by (x − a), remainder = p(a).
Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.

Watch the sign! Dividing by (x + 3) means a = −3. So remainder = p(−3).

Identity Mastery Checklist

See This PatternUse This Identity
a² + 2ab + b²= (a + b)²
a² − 2ab + b²= (a − b)²
a² − b²= (a + b)(a − b)
a³ + b³= (a + b)(a² − ab + b²)
a³ − b³= (a − b)(a² + ab + b²)
a + b + c = 0⇒ a³ + b³ + c³ = 3abc

Quick Self-Check

  1. Degree of 5x³ − 2x + 1? (3)
  2. Remainder when x² + 3x + 2 is divided by (x + 1)? (p(−1) = 1 − 3 + 2 = 0)
  3. Expand: (2a + 3b)² (= 4a² + 12ab + 9b²)
  4. Factorise: 8x³ − 27 (= (2x − 3)(4x² + 6x + 9))

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