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.

Suppose $ a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, ($=y$, say) 
So $ a\sqrt[3]{2} + b\sqrt[3]{4} = y$ and 
$ a\sqrt[3]{2} = y - b\sqrt[3]{4}$
Cubing both sides, we have 
$$ (a\sqrt[3]{2})^3 = (y - b\sqrt[3]{4})^3$$
$$ 2a^3=y^3-4b^3 + 3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}$$
Equating rational and irrational parts, we have $$3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}=0$$  or,
$$3y \cdot 4^{\frac{1}{3}}\left[4^{\frac{1}{3}} - y\right] = 0$$
Now our first assumption that $y$ is rational holds iff $y=0$
But then $\frac{a}{b}=-\frac{\sqrt[3]{4}}{\sqrt[3]{2}}$ implies $a,b$ irrational.
CONTRADICTION.
