proving $\sqrt{a}+\sqrt[3]{b}$ where $a$ and $b$ are integers is rational only if $a$ is a perfect square and $b$ is a perfect cube.

Question

This question was in an oral entry exam for a university in Italy. I have tried to procede by contradiction but don't seem to get anywhere. How can a question like this one be approached?

Answer 1

Assuming $a,\,b\in\Bbb N$:

Let $q:=\sqrt{a}+\sqrt[3]{b}$ so $\sqrt[3]{b}$ is a root of both $x^2-2qx+q^2-a$ and $x^3-b$, so also of$$x^3-b-(x+2q)(x^2-2qx+q^2-a)=(a+3q^2)x+2qa-2a^3-b.$$So if $q\in\Bbb Q$, $\sqrt[3]{b}\in\Bbb Q$ and $b$ is a perfect cube, and $\sqrt{a}=q-\sqrt[3]{b}\in\Bbb Q$ so $a$ is a perfect square.