Degree of the extension $\mathbb Q\subseteq \mathbb Q(\sqrt{5},\sqrt[3]{2})$

Question

Using the tower formula I got that the degree of the extension $\mathbb{Q}\subseteq \mathbb Q(\sqrt{5},\sqrt[3]{2})$ must be a multiple of $6$. I know that the answer is $6$ and I have been trying to find an argument to prove that $$[\mathbb Q(\sqrt{5},\sqrt[3]{2}):\mathbb Q]\leq 6$$ without succeeding. How can I prove that?

Answer 1

We have $$ [\Bbb Q(\sqrt5,\sqrt[3]2):\Bbb Q]=[\Bbb Q(\sqrt5,\sqrt[3]2):\Bbb Q(\sqrt5)]\cdot [\Bbb Q(\sqrt5):\Bbb Q] $$ And from this, $[\Bbb Q(\sqrt5):\Bbb Q]\leq2$ because $\sqrt5$ is a root of $x^2-5$, and $[\Bbb Q(\sqrt5,\sqrt[3]2):\Bbb Q(\sqrt5)]\leq3$ because $\sqrt[3]2$ is a root of $x^3-2$. We see that their product is at most $6$.