Italian Mathematical Olympiad 1992 - Problem 6

Problem:
Let $a$ and $b$ be integers. Prove that if $\sqrt[3]{a}+\sqrt[3]{b}$ is a non-zero rational number, then both $a$ and $b$ are perfect cubes.

Solution:
Let $\sqrt[3]{a}+\sqrt[3]{b}=q \in \mathbb{Q}^*$. From the hypotheses, $a$ and $b$ can't be both zero: without loss of generality we can suppose that $a \neq 0$. Seeking a contradiction, assume that $a$ is not a perfect cube. By Gauss' Lemma, the polynomial $x^3-a$ is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{Z}$: in fact if it was reducible over $\mathbb{Q}$, there would exist $\alpha \in \mathbb{Z}$ such that $\alpha^3 - a = 0$, which contradicts our assumption. Therefore $x^3-a$ is the minimum polynomial of $\sqrt[3]{a}$ over $\mathbb{Q}$ and $[\mathbb{Q}(\sqrt[3]{a}):\mathbb{Q}]=3$. Then, from the initial equation we obtain
$$\sqrt[3]{b}=q-\sqrt[3]{a} \iff (a+b-q^3)+3q^2\sqrt[3]{a} -3q\sqrt[3]{a^2} = 0$$ and since
$\{1,\sqrt[3]{a}, \sqrt[3]{a^2}\}$ is a basis over $\mathbb{Q}$, it must be $a+b=q=0$, which contradicts the hypotheses.