TITLE: Show that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational
QUESTION [6 upvotes]: Our professor asked us this to prove that
$$
\sqrt[3]{2} + \sqrt[3]{4} \notin \Bbb Q.
$$
I know how to prove each one separately that it is irrational, but when it comes to summing two irrational numbers its not certain what the result will be. 
How do I solve it in a more certain way?

REPLY [2 votes]: Write $\sqrt[3]{2} + \sqrt[3]{4} = \alpha + \alpha^2$, with $\alpha=\sqrt[3]{2}$.
Suppose $\alpha + \alpha^2 = q \in \Bbb Q$. Then $\alpha^2 + \alpha^3 = q \alpha$. Since $\alpha^3=2$, we get
$\quad\alpha + \alpha^2 = q$
$\quad\alpha^2 + 2= q \alpha$
Subtracting these, we get $\alpha-2=q-q \alpha$, and so $\alpha=\frac{q+2}{q+1}\in \Bbb Q$, a contradiction since $\alpha$ is irrational.
This argument is essentially dividing the polynomial $x^3-2$ by $x^2+x-q$, getting remainder $(q+1)x-(q+2)$.