Question

Write the following cubes in expanded form:

(i) (2x + 1)3 (ii) (2a – 3b) 3 (iii) [$\frac{3}{2}$ x + 1]3 (iv) [x - $\frac{2 }{ 3}$y]3

Answer 1

(i) It is known that,

(a + b)3 = a3 + b3 + 3ab(a + b) and (a - b)3 = a3 - b3 - 3ab(a - b)

(2x + 1)3 = (2x)3 (1)3 + 3(2x)(1)(2x + 1) = 8x3 + 1 + 6x (2x + 1)

= 8x3 + 1 + 12x2 + 6x = 8x3 + 12x2 + 6x + 1

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

= 8a3 - 27b3 - 18ab (2a - 3b) = 8a3 - 27b3 - 36a2b + 54ab2

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(iii) [$\frac{3 }{ 2}$ x + 1]3 = [$\frac{3 }{ 2}$ x]3 + (1)3 + 3($\frac{3 }{ 2}$x)(1)($\frac{3 }{ 2}$ x + 1)

= $\frac{27 }{ 8}$ x3 + 1 + $\frac{9 }{ 2}$($\frac{3 }{ 2}$ x + 1)

= $\frac{27 }{ 8}$ x3 + 1 + $\frac{27 }{ 4}$x2 + $\frac{9 }{ 2}$2x

= $\frac{27 }{ 8}$ x3 + $\frac{27 }{ 4}$ x2 + $\frac{9 }{ 2}$ x + 1

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(iv) [x - $\frac{2 }{3}$ y]3 = x3 - ($\frac{2 }{3}$ y)3 - 3 (x) ($\frac{2 }{3}$ y)(x - $\frac{2 }{3}$ y)

= x3 - $\frac{8 }{ 27}$ y3 - 2xy (x - $\frac{2 }{ 3}$ y)

= x3 - $\frac{8 }{ 27}$y3 - 2x2 y + $\frac{4 }{ 3}$ xy2.