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.