Is $\mathbb{Q}[α]=\{a+bα+cα^2 :a,b,c ∈ \mathbb{Q}\}$ with $α=\sqrt[3]{2}$ a field? I'm making some exercises to prepare for my ring theory exam:

Is $\mathbb{Q}[α]=\{a+bα+cα^2 :a,b,c ∈ \mathbb{Q}\}$ with
  $α=\sqrt[3]{2}$ a field ?

If $(a+bα+cα^2)(a'+b'α+c'α^2)=1$, then (after quite some calculation and noticing that $α^3=2$ and $α^4=2α$):
\begin{align*}
aa'+2bc'+2cb'&=1 \\ 
ab'+ ba'+2cc'&=0 \\
ca'+bb'+ac' &= 0
\end{align*}
I'm not sure how to proceed, and if I'm heading in the right direction. Any help would be appreciated.

Something else I was thinking about, this ring I have seems to be isomorphic to:
$$\mathbb{Q}[X]/(X^3-2)$$
But this is not a maximal ideal, as it is contained in the ideal $(X^3,2)$. Would this be correct reasoning ?
 A: One approach would be to consider  $\mathbb{Q}[\alpha]$ as a $\mathbb{Q}$-vector space of dimension $3$.
Let $\beta = a + b\alpha + c\alpha^2$.  Now consider the numbers $1$, $\beta$, $\beta^2$, and $\beta^3$.  Because $\mathbb{Q}[\alpha]$ has dimension $3$, these are linearly dependent, so there are $u,v,w,x\in \mathbb{Q}$ with $u+v\beta + w\beta^2 + x\beta^3 =0$.
We can assume $u\neq 0$ (why?), so $\beta(v+w\beta+x\beta^2)=-u$, and we're almost done...
A: There is a purely abstract way to do this in higher generality:
Let $K$ be a field and suppose $\alpha$ exists in an extension and is algebraic over $K$. Therefore we can say that $K[\alpha]\cong K[x]/f(x)$ where $f(x)$ is $\alpha$'s minimal polynomial over $K$. We want to show that any nonzero element $g(x)\in K[x]/f(x)$ is invertible, i.e. $\exists u,v$ s.t. $u(x)f(x)+v(x)g(x)=1$...
A: There is also a low-brow approach.
It will suffice that the three equations you wrote out always have a solution $a',b',c'$ for any $a,b,c$, except $0,0,0$.
This is a linear system of equations, with the corresponding matrix $$A = \begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a 
\end{bmatrix}.$$
So, to check that there is a solution, you just need to check that $\det A \neq 0$. But $\det A = a^3 + 2b^3 + 4 c^3 - 6abc$, which is non-negative by the inequality between arithmetic and geometric means, and which would be $0$ only if $a^3 = 2b^3 = 4c^3$ - but this cannot happen for non-zero integers. 
