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x^3 + y^3 + z^3 - 3xyz

Find the cube...

Question

Find the cube of the following binomial expression:

(iii) $2 x + \frac{3}{x}$

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Solution

Given, $2 x + \frac{3}{x}$

Using Identity,
$(a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)$

$∴ {(2 x + \frac{3}{x})}^{3}$

$= (2 x)^{3} + {(\frac{3}{x})}^{3} + 3 (2 x \times \frac{3}{x}) (2 x + \frac{3}{x})$

$= 8 x^{3} + \frac{27}{x^{3}} + 18 (2 x + \frac{3}{x})$

$= 8 x^{3} + \frac{27}{x^{3}} + 18 \times 2 x + 18 \times \frac{3}{x}$

$= 8 x^{3} + \frac{27}{x^{3}} + 36 x + \frac{54}{x}$

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