Question

Rationalise the denominator of $\frac{1}{\sqrt[3]{32+1}}$.

Answer 1

The given term is $\frac{1}{\sqrt[3]{32+1}}=\frac{1}{\sqrt[3]{33}}$.

Rationalize th denominator two times,

$\frac{1}{\sqrt[3]{33}}=\frac{1}{\sqrt[3]{33}}\times \frac{\sqrt[3]{33}}{\sqrt[3]{33}}\times \frac{\sqrt[3]{33}}{\sqrt[3]{33}}$

$=\frac{\sqrt[3]{33^2}}{\sqrt[3]{33^3}}$

$=\frac{\sqrt[3]{39}}{3}$

$=\frac{\sqrt[3]{39}}{3}$