Prove that $\mathbb{Q}(\sqrt[3]{p}, \sqrt{q}) = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})$

Question

Let $p, q$ be distinct primes and $L := \mathbb{Q}(\sqrt[3]{p}, \sqrt{q})$.

I want to show that $L = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})$. Of course one inclusion is clear, but how do I get the other one?

Answer 1

$\sqrt[3]p = \frac {(\sqrt[3]p\cdot \sqrt q)^4}{pq^2}$

$\sqrt q = \frac {(\sqrt[3]p\cdot\sqrt q)^3}{pq}$

Answer 2

You want to see that $\sqrt{q}\in\mathbb{Q}(\sqrt[3]{p}\sqrt{q})$.

Hint: cube.