Question

Find the cube of the following binomial expression:

(i) $\frac{1}{x}+\frac{y}{3}$

Answer 1

Given, $\frac{1}{x}+\frac{y}{3}$

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

$\therefore {(\frac{1}{x}+\frac{y}{3})}^3={\left(\frac{1}{x}\right)}^3+{\left(\frac{y}{3}\right)}^3+3(\frac{1}{x}\times \frac{y}{3})(\frac{1}{x}+\frac{y}{3})$

$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x}(\frac{1}{x}+\frac{y}{3})$

$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x}\times \frac{1}{x}+\frac{y}{x}\times \frac{y}{3}$

$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x^2}+\frac{y^2}{3x}$