Question

Rationalise the denominator of $\frac{1}{2^{\frac{1}{3}}+2^{\frac{-1}{3}}}$

Answer 1

Given, $2^{\frac{1}{3}}+2^{\frac{-1}{3}}=2^{\frac{1}{3}}+\frac{1}{2^{\frac{1}{3}}}$ $=\sqrt[3]{32}+\sqrt[3]{3\frac{1}{2}}$

Its rationalising factor is $\sqrt[3]{32^2}-\sqrt[3]{32\times \frac{1}{2}}+\sqrt[3]{3{\left(\frac{1}{2}\right)}^2}$

$=\sqrt[3]{32^2}-\sqrt[3]{31}+\sqrt[3]{3{\left(\frac{1}{2}\right)}^2}$

$=\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}$

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

$=\frac{1}{\sqrt[3]{32}+\sqrt[3]{3\frac{1}{2}}}\times \frac{\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}}{\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}}$

$=\frac{\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}}{(2^{\frac{1}{3}}+2^{\frac{-1}{3}})(\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}})}$

$=\frac{\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}}{{\left(\sqrt[3]{32}\right)}^3+{\left(\sqrt[3]{3\frac{1}{2}}\right)}^3}$

$=\frac{\sqrt[3]{34}-1+\sqrt[3]{3\frac{1}{4}}}{2+\frac{1}{2}}$

$=\frac{\sqrt[3]{32^2}-1+\sqrt[3]{3\frac{1}{2^2}}}{\frac{5}{2}}$

$=\frac{2}{5}(2^{\frac{2}{3}}+2^{\frac{-2}{3}}-1)$