What is $\frac{1}{1+\sqrt[3]{2}}$ in $\mathbb{Q}(\sqrt[3]{2})$?

Use the expansion $(1+x^3)=\left( x+1 \right) \left( {x}^{2}-x+1 \right)$. Then

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

Here is a way of approaching this problem in general.

In $\mathbb Q(i)/\mathbb Q$ or more generally in $\mathbb C/\mathbb R$, in order to "rationalise the denominator" of $$\frac{1}{a+bi}$$we multiply the top and bottom of the fraction by the complex conjugate $$\frac{1}{a+bi}=\frac{1}{a+bi}\cdot\frac{a-bi}{a-bi}=\frac{a-bi}{a^2+b^2}.$$

So we're looking for some kind of generalisation of the complex conjugation map in a more general field. Complex conjugation has two important properties that we will want to preserve:

1. If $a\in\mathbb R$ then $\overline a = a$. So complex conjugation acts on $\mathbb C$, preserving $\mathbb R$.
2. For any $a\in \mathbb C$, $a\overline a \in\mathbb R$, so by multiplying by the complex conjugate, we guarantee that the denominator will be real.

The crucial reason that complex conjugation has these properties is this: there is an abstract field $$K=\mathbb R[X]/(X^2+1)$$which we would usually write as $\mathbb C$. But there are two natural ways that we can view $K$ as $\mathbb C$; two natural embeddings of $K$ into $\mathbb C$:

1. $K\to \mathbb C$, $X\mapsto i$,
2. $K\to \mathbb C$, $X\mapsto -i$.

The point is that our choice of which root of $X^2+1$ is "$i$" and which is "$-i$" is arbitrary. Both are equally valid. The product $i\overline i$ must be $1$ because $1$ is the constant term of the polynomial $X^2+1$. Since complex conjugation preserves $\mathbb R$, this shows that $a\overline a \in \mathbb R$ for any $a\in\mathbb C$.

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So how can we generalise this? As an abstract field, we have $$K=\mathbb Q[X]/(X^3-2).$$ One way of embedding this into $\mathbb C$ is to send $X\mapsto \sqrt[3]2$, and doing this gives the field $\mathbb Q(\sqrt[3]2)$. But there are two other equally valid embeddings into $\mathbb C$ given by sending $X$ to the other roots of $X^3-2$ $$\sigma:X\mapsto\zeta\sqrt[3]2\\\overline\sigma:X\mapsto\zeta^2\sqrt[3]2$$where $\zeta$ is a primitive cubed root of unity, and satisfies $\zeta^2+\zeta+1=0$. As before, $\sigma$ and $\overline\sigma$ act trivially on $\mathbb Q$, and you can check that if $a\in\mathbb Q(\sqrt[3]2)$ the product of the embeddings (which is called the norm of the extension) $$a\cdot\sigma(a)\cdot\overline\sigma(a)\in\mathbb Q.$$

For a general finite extension of fields $L/K$, and an element $x\in L$, the norm of $x$ can be defined by $$N_{L/K}(x)=\prod_{\sigma\in\rm{Aut}_K(L)}\sigma(x)$$where the product is over the automorphisms of $L$ which preserve $K$ (equivalently, the embeddings of $L$ into some algebraic closure of $K$). It is a general fact (which uses the ideas above) that for any $x\in L$, we have $N_{L/K}(x)\in K$.

So analogously to before, $$\begin{align} \frac{1}{1+\sqrt[3]2}&=\frac{1}{1+\sqrt[3]2}\cdot\frac{(1+\zeta\sqrt[3]2)(1+\zeta^2\sqrt[3]2)}{(1+\zeta\sqrt[3]2)(1+\zeta^2\sqrt[3]2)}\\&=\frac{1}{1+\sqrt[3]2}\cdot\frac{1-\sqrt[3]2+\sqrt[3]4}{1-\sqrt[3]2+\sqrt[3]4}\\&=\frac{1-\sqrt[3]2+\sqrt[3]4}{3} \end{align}$$

You should note the similarity of this method in the end to the other answer.

Can you see what to do for a general element of $\mathbb Q(\sqrt[3]2)$? You should deduce that $$\frac{1}{a+b\sqrt[3]2+c\sqrt[3]4}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]2+(b^2-ac)\sqrt[3]4}{a^3+2b^3+4c^3-6abc}$$