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.

Really you are saying: if $\alpha,\beta\in\Bbb Q(\sqrt{2})$ and $\beta\neq0$, then $\alpha/\beta$ can be expressed in terms of the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ i.e. $\alpha/\beta\in\Bbb Q(\sqrt[3]{2})$. Written like so, the answer is obvious since $\Bbb Q(\sqrt[3]{2})$ is a field... the key point being: $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is actually a basis. So why is that true?
I'll make an abstract argument here that you can massively generalise to any algebraic field extension. $\Bbb Q(\sqrt[3]{2})\cong\Bbb Q[x]/(x^3-2)$ where the isomorphism symbolically evaluates each polynomial class on the RHS at $x=\sqrt[3]{2}$ (this is creating the linear span of $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$). Since $\Bbb Q$ is a field, $\Bbb Q[x]$ is a principal ideal domain; since $x^3-2$ is an irreducible polynomial, the ideal $(x^3-2)$ is maximal - if another ideal $J$ contains it, then $J=(p)$ for some polynomial $p$ and $(p)\supset(x^3-2)$ means $p$ divides $(x^3-2)$, and the only possible divisors $p$ are $(\pm)(x^3-2)$ or $(\pm)1$ so $J$ is always either $(x^3-2)$ itself or the entire ring $\Bbb Q[x]$. Maximality means $\Bbb Q[x]/(x^3-2)$ is a field, because: if a nonzero element $p$ is not invertible, then $(p)$ does not contain $1$, so $(p)\neq\Bbb Q[x]/(x^3-2)$ and it is also a nonzero ideal, which means $(p)$ is a proper container of $(x^3-2)$ when lifted back into $\Bbb Q[x]$ - this is impossible. Therefore all nonzero elements are invertible!
What on Earth does all that mean? Suppose an element $p=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is not invertible (there do not exist $\alpha,\beta,\gamma\in\Bbb Q$, $(\alpha+\beta\sqrt[3]{2}+\gamma\sqrt[3]{4})p=1$). Then the set of all multiplications $p\cdot(\alpha+\beta\sqrt[3]{2}+\gamma\sqrt[3]{4})$ does not contain $1$ so is not the entirety of $\Bbb Q(\sqrt[3]{2})$, but if $p\neq0$ then this set contains nonzero elements (namely, $p$...) - let's call this set (principal ideal) $J_1$. Consider all polynomials $a_0+a_1x+a_2x^2+\cdots+a_nx^n$ in rational coefficients. For any $q\in J_1$ and some polynomial $f_0$, we have that $q=f_0(\sqrt[3]{2})$. In fact, this is true for very many $f_0$ - any $f(x)=f_0(x)+g(x)(x^3-2)$ will have $f(\sqrt[3]{2})=p$. Let's call the set of all such $f(x)$ (ranging over $q\in J_1$) and all multiplications $f(x)h(x)$ the set (principal ideal) $J_2$, where $h(x)$ is any rational polynomial. Our assumptions are that $J_2\neq\{0\}$ and $J_2\neq\Bbb Q[x]$ - $1\notin J_2$. Moreover, since $q=0$ is an option, so $f_0(x)=0$ is an option and thus $f(x)=x^3-2$ is an option, we know that $x^3-2\in J_2$ as well as all multiples of $x^3-2$.
But it follows from polynomial division that some polynomial $T(x)$ must divide every element of $J_2$. That means $T(x)$ divides $x^3-2$, so either $T=\pm1$ or $T=\pm(x^3-2)$. In the first case, we get $J_2=\Bbb Q[x]$ which is false. In the second case, we get $J_2$ to be all multiples of $x^3-2$. That means, when we evaluate any element of $J_2$ at $\sqrt[3]{2}$, we get zero. In particular it means $p=0$. But this is a contradiction! (the same as the one above, but hopefully more concrete). It follows that there must be some $\alpha+\beta\sqrt[3]{2}+\gamma\sqrt[3]{4}$ that serves as an inverse to $p$.
Even more concretely: (following Will Jagy's hint):

$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(a^2+b^2\sqrt[3]{4}+c^2\sqrt[3]{16}-bc\sqrt[3]{8}-ac\sqrt[3]{4}-ab\sqrt[3]{2})=a^3+b^3(2)+c^3(4)-3abc\sqrt[3]{8}$. In other words: $$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(a^2-2bc+[2c^2-ab]\sqrt[3]{2}+[b^2-ac]\sqrt[3]{4})=a^3+2b^3+4c^3-6abc$$So: $$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}$$

