How to prove that $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$?

Question

It's clear that $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})\subseteq\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$, but for the other direction, the most simple thing I could think of is to calculate powers of $\sqrt[3]{2}+\sqrt{3}$ (let's call it $\alpha$), and after some work with matrices, I got $-1638+3204\alpha-468\alpha^{2}-720\alpha^{3}+18\alpha^{4}+72\alpha^{5}=1836\sqrt{3}$ (and hence $\sqrt{3},\sqrt[3]{2}\in\mathbb{Q}(\alpha)$.

I've a feeling that there is a much simpler way to show that. What am I missing?

Answer 1

Set $\alpha=\sqrt[3]{2}+\sqrt{3}$, so that $\alpha-\sqrt{3}=\sqrt[3]{2}$ and, cubing, $$ \alpha^3-3\alpha^2\sqrt{3}+9\alpha+3\sqrt{3}=2 $$ Therefore $\sqrt{3}\in\mathbb{Q}(\alpha)$.