Prove that $\mathbb Q[\sqrt[3]2]$ is a field We define the set: 
$$\mathbb{Q}[\sqrt[3]2]=\{a_{0}+a_{1}\sqrt[3]{2}+a_{2}\sqrt[3]{2^{2}}:a_{0}, a_1,a_2\in\mathbb{Q}\}$$
It's easy to prove all the properties of fields, except for the unit elements.
So, how can we prove that
$$\forall x\in\mathbb{Q^{*}}[\sqrt[3]2],\exists x^{-1} \in\mathbb{Q}[\sqrt[3]2]:xx^{-1}=1$$ 
And how can we prove this in general for the set
$$\mathbb{Q}[\sqrt[n]2]=\{a_{0}+a_{1}\sqrt[n]{2}+a_{2}\sqrt[n]{2^{2}}+...+a_{n-1}\sqrt[n]{2^{n-1}}:a_{0}, a_1,a_2,...,a_{n-1}\in\mathbb{Q}\}$$
 A: It's quite fiddly to do this directly, especially in general, although it is possible. Effectively, you have to come up with a generalisation of the idea of "rationalising the denominator".  It turns out that $$\frac{1}{a+b\sqrt[3]2+c\sqrt[3]4}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]2+(b^2-ac)\sqrt[3]4}{a^3+2b^3+4c^3-6abc}.$$ Clearly, this isn't something you want to work out from scratch, and I don't fancy trying to generalise this. 
As is often the case, it is better is to work abstractly.
First show that $\mathbb Q[\sqrt[3]2]\cong\mathbb Q[X]/(X^3-2)$. To show the latter is a field, all we need to do is show that $(X^3-2)$ is a maximal ideal of $\mathbb Q[X]$. Since $\mathbb Q[X]$ is a PID, this is equivalent to showing that the polynomial $X^3-2$ is irreducible. 
A: In the case of the cubic ring $\Bbb Q[\lambda]$ where $\lambda^3=2$, it’s really quite easy to find the reciprocal of $z=a+b\lambda+c\lambda^2$ by the method of rationalizing the denominator. You use the fact that this quantity has the two conjugates $z'=a+b\omega\lambda+c\omega^2\lambda^2$ and $z''=a+b\omega^2\lambda+c\omega\lambda^2$, where $\omega$ is a primitive cube root of unity, $\omega^2+\omega+1=0$.
Just as you expect, you get $$\frac1{z}=\frac{z'z''}{zz'z''}\,,$$
where the numerator is in $\Bbb Q[\lambda]$ and the denominator is in $\Bbb Q$: the resulting formula is exactly the one that @Mathmo123 has given. Since I have done exactly this computation purely by hand many times when I was in graduate school, I do not consider it at all fiddly, but rather pleasingly instructive.
In general, I recommend strongly that you not shrink from hand computation.
A: Doing this for $\sqrt[3]{2}$ is a waste of time. ;-) Exactly because doing it for $\sqrt[n]{2}$ would lead to gigantic computations.
Suppose $r\in\mathbb{C}$ is algebraic over $\mathbb{Q}$. We want to see that the set $\mathbb{Q}[r]$ consisting of all the expressions of the form $a_0+a_1r+\dots+a_nr^n$ is a field.
Let $r$ be an algebraic element over the field $F$. Then $F[r]$ is precisely the image of the ring homomorphism $\varphi\colon F[X]\to F[r]$ which is the identity on $F$ and $\varphi(X)=r$.
By the homomorphism theorem,
$$
F[r]\cong F[X]/\ker\varphi
$$
Now, if $f(X)$ is the minimal polynomial for $r$ over $F$, we can easily see that $\ker\varphi=(f(X))$, the principal ideal generated by $f(X)$.
As $f(X)$ is irreducible, $(f(X))$ is a maximal ideal, so $F[X]/(f(X))$ is a field.

For the particular case, consider the map $\mathbb{Q}[\sqrt[3]{2}]\to\mathbb{Q}[\sqrt[3]{2}]$ given by $t\mapsto t(a+b\sqrt[3]{2}+c\sqrt[3]{4}\,)$. It is a $\mathbb{Q}$-linear map and, with respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, its matrix is
$$
\begin{bmatrix}
a & 2c & 2b \\
b & a & 2c \\
c & b & a
\end{bmatrix}
$$
whose inverse (one of the few cases where using the adjugate is simpler than other methods) is
$$
\frac{1}{a^3+2b^3+4c^3-6abc}
\begin{bmatrix}
a^2 - 2bc & -2ac + 2b^2 & -2ab + 4c^2 \\
-ab + 2c^2 & a^2 - 2bc & -2ac + 2b^2 \\
-ac + b^2 & -ab + 2c^2 & a^2 - 2bc
\end{bmatrix}
$$
which, by the way, proves that $a^3+2b^3+4c^3-6abc\ne0$ as soon as one among $a$, $b$ and $c$ is nonzero.
The inverse of $a+b\sqrt[3]{2}+c\sqrt[3]{4}\ne0$ is thus
$$
\frac{(a^2-bc)+(-ab+2c^2)\sqrt[3]{2}+(-ac+b^2)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}
$$
