Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to see the motivation behind this.  Can someone give a more general explanation of how to come to this expression?

You rationalize by using Euclid's algorithm as used to prove Bézout's identity. 
Let $\alpha = \sqrt[3]{2}$. We have obviously to assume $a+b\sqrt[3]2 + c(\sqrt[3]2)^2 = a+b\alpha + c \alpha^2 \ne 0$.
Consider the polynomial $0 \ne a + b x + c x^2 \in \mathbb{Q}[x]$. Since $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$, for the gcd we have $(a + b x + c x^2, x^3 - 2) = 1$. Use Euclid's algorithm to find polynomials $u, v \in \mathbb{Q}[x]$ such that $(a + b x + c x^2) \cdot u + (x^3 - 2) \cdot v = 1$. Now evaluate this for $x = \alpha$ to obtain $(a + b \alpha+ c\alpha^2) \cdot u(\alpha) = 1$, so that
\begin{equation}
\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}
=
\frac{1}{a + b \alpha+ c \alpha^2}
=
u(\alpha).
\end{equation}
