So we want to find an $u$ such that $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$. I obtained that if $u$ is of the following form: $u=\sqrt[6]{2^a5^b}$Where $a\equiv 1\pmod{2}$, and $a\equiv 0\pmod{3}$, and $b\equiv 0\pmod{2}$ and $ b\equiv 1\pmod{3}$. This works since $u^3=\sqrt{2^a5^b}=2^{\frac{a-1}{2}}5^{\frac{b}{2}}\sqrt{2}$and also, $u^2=\sqrt[3]{2^a5^b}=2^{\frac{a}{3}}5^{\frac{b-1}{3}}\sqrt[3]{5}$Thus we have that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq \mathbb{Q}(u)$. Note that $\sqrt{2}$ has degree of $2$ (i.e., $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$) and alsothat $\sqrt[3]{5}$ has degree $3$. As $\gcd(2,3)=1$, we have that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{5}),\mathbb{Q}]=6$. Note that this is also the degree of the extension of $u$, since one could check that the set $\{1,u,...,u^5\}$ is $\mathbb{Q}$-independent. Ergo, we must have equality. That is, $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$.
My question is: How can I find all such $w$ such that $\mathbb{Q}(w)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$? This is homework so I would rather hints rather hints than a spoiler answer. I believe that They are all of the form described above, but apriori I do not know how to prove this is true.
My idea was the following, since $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ has degree $6$, then if $w$ is such that the desired equality is satisfied, then $w$ is a root of an irreducible polynomial of degree $6$, moreover, we ought to be able to find rational numbers so that $\sqrt{2}=\sum_{i=0}^5q_iw^i$ and $\sqrt[3]{5}=\sum_{i=0}^5p_iw^i$But from here I do not know how to show that the $u$'s described above are the only ones with this property (It might be false, apriori I dont really know).