How am I supposed to prove that this fraction is irrational?

Question

$$\frac{\sqrt2}{1+\sqrt[3]2}\notin\Bbb Q.$$

I have proved that both numbers are irrational but I don't think that is sufficient to prove that the fraction is irrational

Answer 1

If $p=\frac{\sqrt{2}}{1+\sqrt[3]{2}}\in \mathbb{Q} $ then $p^2 =\frac{2}{1+2\sqrt[3]{2}+\sqrt[3]{4}}\in\mathbb{Q}$ and also $\frac{2}{p^2}-1=2\sqrt[3]{2}+\sqrt[3]{4}\in\mathbb{Q}$ but then $\left(\frac{2}{p^2}-1\right)^2=4\sqrt[3]{4}+8 +2\sqrt[3]2 \in\mathbb{Q}$ and then $$\sqrt[3]{4}=\frac{\left(\frac{2}{p^2}-1\right)^2-\left(\frac{2}{p^2}-1\right) -8}{3}\in \mathbb{Q}$$

but the last is impossible.