Rationalize the denominator:
$\displaystyle\frac{\sqrt[3]{2}}{\sqrt[3]{2}+\sqrt[3]{3}}$.
Solution. We
use the fact that
a^3+b^3 = (a+b)(a^2-ab+b^2).
Let $a=\sqrt[3]{2}$ and $b=\sqrt[3]{3}$.
Then $a^3+b^3=5$ and we write
$\begin{eqnarray}
\frac{\sqrt[3]{2}}{\sqrt[3]{2}+\sqrt[3]{3}} &=&
\frac{\sqrt[3]{2}}{\sqrt[3]{2}+\sqrt[3]{3}}\cdot \frac{(\sqrt[3]{2})^2
- \sqrt[3]{2}\sqrt[3]{3} + (\sqrt[3]{3})^2}{(\sqrt[3]{2})^2 -
\sqrt[3]{2}\sqrt[3]{3} + (\sqrt[3]{3})^2} \\
&=& \frac{\sqrt[3]{2} (\sqrt[3]{4} -
\sqrt[3]{6}+\sqrt[3]{9})}{5} \\
&=& \frac{\sqrt[3]{6} - \sqrt[3]{12} + \sqrt[3]{18}}{5}.
\end{eqnarray}$