<s>[INST] 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!
 [/INST] $y = \sqrt[3]{2} + \sqrt[3]{4}$ is a root of the equation $y^3 - 6y - 6 = 0$.
(To see this, let $x = \sqrt[3]{2}$ and $y = \sqrt[3]{2}+\sqrt[3]{4}=x+x^2$. Then $y^3 = x^3 + 3x^4 + 3x^5 + x^6 = 2 + 6x + 6x^2 + 4 = 6 + 6y$.)
By the rational root theorem, we know that any rational roots of $y^3 - 6y - 6 = 0$ would have to be in the set $\{\pm1,\pm2,\pm3,\pm6\}$; we can quickly confirm that none of these is in fact a root, meaning that the equation does not have any rational roots.
Therefore, $y$ must be irrational.
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