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 ?

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. 
