Showing $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$.

Question

Let $p$ and $q$ be prime numbers, $p\neq q$.

Now I have to show $L:=\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})=:K$.

The direction $K\subseteq L$ is clear. For the other direction I have to show that $\sqrt{p}\in K$ and $\sqrt[3]{q}\in K$. Thats where I'm at at the moment.

Can someone please give me a hint on how to show this?

Answer 1

Hint $(\sqrt p \sqrt[3]{q})^3=qp\sqrt p$