TITLE: Let $a=\sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5} \in \mathbb{R}$. Show that $a$ is algebraic over $\mathbb{Q}$ and that $|\mathbb{Q}(a):\mathbb{Q}|\le 30$.
QUESTION [5 upvotes]: I know that if $a$ is algebraic over $\mathbb{Q}$ then there exists minimal polynomial $f(x)\in Z[x]$ such that $a$ is the root of that polynomial. I am trying to find that polynomial which I know will be of degree $30$ so that $|\mathbb{Q}(a):\mathbb{Q}|\le 30$.
I have looked at other examples on this forum like when $a=\sqrt{3}+\sqrt{5}$ or $a=\sqrt[3]{2}$ or $a=\sqrt[3]{3}+\sqrt[3]{9}$ and looked at their methods of finding the required polynomial but each method is different and the given $a=\sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5}$, it is difficult to find its polynomial using any their methods.
If I try solving it following way $(a-\sqrt{2})^2=(\sqrt[3]{3}+\sqrt[5]{5})^2 \Rightarrow a^2+2-2a\sqrt{2}=\sqrt[3]{9}+\sqrt[5]{25}-2\sqrt[3]{3}\sqrt[5]{5}$ and then go on from there, then I get a messy equation which can't be solved to get a polynomial $f$ in $a$ for which $f(a)=0$. Any help would be appreciated.

REPLY [2 votes]: Hint: $a \in \mathbb Q(\sqrt{2},\sqrt[3]{3},\sqrt[5]{5})$ which is a finite extension of $\mathbb Q$ of degree at most $30$:
$$
[\mathbb Q(u,v,w):\mathbb Q]=[\mathbb Q(u,v,w):\mathbb Q(u,v)][\mathbb Q(u,v):\mathbb Q(u)][\mathbb Q(u):\mathbb Q] \le 5 \cdot 3 \cdot 2 = 30
$$
where $u=\sqrt{2}$, $v=\sqrt[3]{3}$, $w=\sqrt[5]{5}$.