<s>[INST] What is $\frac{1}{1+\sqrt[3]{2}}$ in $\mathbb{Q}(\sqrt[3]{2})$? Since $\mathbb{Q}(\sqrt[3]{2})$ is a field, any number $\neq 0$ has a reciprocal.  
How then to write $\frac{1}{1+\sqrt[3]{2}}$ as a number $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ with fractions $a,b,c \in \mathbb{Q}$?

In general, how to "rationalize denominators" in a cubic field such as $\mathbb{Q}(\sqrt[3]{2})$ ?
$$ \frac{1}{a + b\sqrt[3]{2} + c\sqrt[3]{4}}$$
 [/INST] Use the expansion $(1+x^3)=\left( x+1 \right)  \left( {x}^{2}-x+1 \right)$.
Then 
$$
\frac{1}{1+\sqrt[3]{2}}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{(1+\sqrt[3]{2})(1-\sqrt[3]{2}+(\sqrt[3]{2})^2)}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{3}=\frac{1}{3}-\frac{\sqrt[3]{2}}{3}+\frac{(\sqrt[3]{2})^2}{3}
$$
</s>