Question

Find the cube of the following binomial expression:

(ii) $\frac{3}{x}-\frac{2}{x^2}$

Answer 1

Given ,$\frac{3}{x}-\frac{2}{x^2}$

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

$\therefore {(\frac{3}{x}-\frac{2}{x^2})}^3={\left(\frac{3}{x}\right)}^3-{\left(\frac{2}{x^2}\right)}^3$

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

$=\frac{27}{x^3}-\frac{8}{x^6}-\frac{18}{x^3}(\frac{3}{x}-\frac{2}{x^2})$

$=\frac{27}{x^3}-\frac{8}{x^6}-\frac{18}{x^3}\left(\frac{3}{x}\right)+\frac{18}{x^3}\left(\frac{2}{x^2}\right)$

$=\frac{27}{x^3}-\frac{8}{x^6}-\frac{54}{x^4}+\frac{36}{x^5}$