TITLE: How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational?
QUESTION [29 upvotes]: 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!

REPLY [5 votes]: Here is another approach which generalises to many similar examples, and which involves no complicated simplifications of surds.  (In fact, no easy simplifications of surds either.)
First, $\sqrt[3]2$ is an algebraic integer, that is, it is a root of
$$x^3-2$$
which is a monic polynomial (leading coefficient $1$) with integer coefficients.  Similarly, $\sqrt[3]4$ is an algebraic integer.
It is known that the sum of two algebraic integers is an algebraic integer.  Thus, $x=\sqrt[3]2+\sqrt[3]4$ is an algebraic integer.
It is also known that if an algebraic integer is rational, then it is a rational integer - that is, an ordinary integer, $0,1,-1,2,-2$ etc.  However, it is not hard to find the estimates
$$1<\sqrt[3]2<\frac43 ,\quad \frac43<\sqrt[3]4<\frac53\ ;$$
so $\frac73<x<3$, hence $x$ is not an integer and must be irrational.