Question

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

Answer 1

$\frac{1}{\sqrt[3]{32+1}}=\frac{1}{2^{\frac{1}{3}}+1}\times \frac{(2^{\frac{1}{3}}{)}^2-2^{\frac{1}{3}}+1}{(2^{\frac{1}{3}}{)}^2-2^{\frac{1}{3}}+1}$

$=\frac{2^{\frac{2}{3}}-2^{\frac{1}{3}}+1}{2+1}$ {$\because a^3+b^3=(a+b)(a^2-ab+b^2)$}

$=\frac{2^{\frac{2}{3}}-2^{\frac{1}{3}}+1}{3}$