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CBSE Class 9 Mathematics Question 20 of 30

Polynomials — Question 6

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6
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