prove that $\dfrac{\alpha}{\beta}$ can be written in the form $a + b\sqrt[3]{2} +c\sqrt[3]{4}$ for rationals a,b,c. 
Let $\alpha = \sum_{i=0}^2 a_i(\sqrt[3]{2})^i$ where each $a_i$ is rational and let $\beta = \sum_{i=0}^2 b_i (\sqrt[3]{2})^i$ where each $b_i\in \mathbb{Q}$ and not all the $b_i$'s are zero. Prove that $\dfrac{\alpha}{\beta}$ can be written in the form $a + b\sqrt[3]{2} +c\sqrt[3]{4}$ for rationals a,b,c.

I'm not sure how to "rationalize" the denominator of $\dfrac{\alpha}{\beta}.$ The standard trick for square roots obviously doesn't work here, but I think there's a variant that might work. I know $1+\sqrt[3]{2}+\sqrt[3]{4} = \dfrac{1}{\sqrt[3]{2}-1}.$ One can of course assume all the $a_i$'s and $b_i$'s are integers by multiplying the numerator and denominator of $\alpha/\beta$ by the least common multiple of the $a_i$'s and $b_i$'s.

Without any theory of number you may proceed as folow:
Your problem is just to find a polynomial form for inverse. The technic is the same as for $\Bbb Z/n\Bbb Z$.
$X^3-2$ and any $a+bX+cX^2$ are relatively prime because no cubic roots of 2 can cancel a polynom in $\Bbb Q[X]$ with degree=2. Then there exist $U,V\in\Bbb Q[X]$ with
$U(X)(a+bX+cX^2)+V(X)(X^3-2)=1$ then substituing $\sqrt[3]2$ to $X$ you get
$$\frac 1{a+b\sqrt[3]2+c(\sqrt[3]2)^2}=U(\sqrt[3]2)$$
