Trivial: Rationalize fraction with a third-degree root This is a pretty trivial question. How do you rationalize a function with a denominator that contains a third degree root?

Edit: My expression is $\displaystyle{\frac{1}{\sqrt[3]{2}-1}}$.

To rationalize when you have a denominator of the form $a^{1/3}-b^{1/3}$ (as you do here, with $a=2$ and $b=1$), use the identity
$x^3-y^3 = (x-y)(x^2+xy+y^2)$.
So
$$2-1 = \left(\sqrt[3]{2}-\sqrt[3]{1}\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\right) = \left(\sqrt[3]{2} - 1\right)\left(\sqrt[3]{4}+\sqrt[3]{2} + 1\right).$$
So, multiply both numerator and denominator by $\sqrt[3]{4}+\sqrt[3]{2}+1$. You get:
$$\begin{align*}
\frac{1}{\sqrt[3]{2}-1} &= \frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}\\
&= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{(\sqrt[3]{2})^3 - 1^3)}\\
&= \frac{\sqrt[3]{4} + \sqrt[3]{2}+1}{2-1}\\
&= \sqrt[3]{4} + \sqrt[3]{2} + 1.
\end{align*}$$
If you had a denominator of the form $a^{1/3}+b^{1/3}$, you can instead use the identity $(x^3+y^3) = (x+y)(x^2-xy+y^2)$.
