Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational? I need to prove that $$ a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b $  are non zero  rationals.
I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational $\implies$ $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, so I could show it is a contradiction.

If $b\ne0$, then $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational iff $\sqrt[3]{2}$ is a root of a quadratic polynomial over $\mathbb Q$.
$\sqrt[3]{2}$ is a root of $X^3-2$, which is irreducible over $\mathbb Q$.
So, every polynomial having $\sqrt[3]{2}$ as root must be a multiple of $X^3-2$. In particular, $\sqrt[3]{2}$ is not a root of a quadratic polynomial over $\mathbb Q$.
Therefore, 
if $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, then $b=0$.
But $\sqrt[3]{2}$ is irrational, and so $a=0$.
All this can be summarized as:
$1, \sqrt[3]{2}, \sqrt[3]{4}$ are linearly independent over $\mathbb Q$ and so the only way that $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational is when $a=b=0$.
