How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:
Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.
I also figured maybe I should play with it this way:
$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$
But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.
So what am I missing (other than sleep and food)? What route should I take to 
prove this?
Thanks in advance!

It holds true for any $ $ irrational $\,x=\sqrt[3]n,\,$ except $\,n=1\,$ (so $\,x^2+x=-1\in\Bbb Q)$
More generally: $ $ if $\ r\in\Bbb Q\ $ and $\,x=\sqrt[3]r\not\in\Bbb Q\,$ then $\,x^2+x = q\in\Bbb Q\iff r = 1.$
Proof $\,\ qx = x^3+x^2 = r+x^2\ $ so $\ qx-r = x^2 = q-x,\ $ so $\,(\color{#c00}{q\!+\!1})\,x = r+q.$
Therefore $\,\ x\not\in\Bbb Q\,\Rightarrow\,\color{#c00}{q = -1}\,\Rightarrow\,  0 = (\color{#c00}{q\!+\!1})x = r+q = r-1,\ $ thus $\,\ r = 1.$ 
