નીચેના વિસ્તરણ કરો : 

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

$\because \quad(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$

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

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

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