Show that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ 
Show that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ and find all $w\in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ such that $\mathbb{Q}(w)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$.

It is clear that  $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}) \subseteq \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$. But given any $x\in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$, how do I show that $x\in \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? For the second question, is $w$ the associates of $\sqrt{2}+\sqrt[3]{5}$? How do I show that? Thanks!

IMO, Galois theory is the way to go here, but I want to give you an idea of how to do this in an elementary way.  The key idea below is that squaring should "eliminate" some of the $\sqrt{2}$, but won't really get rid of any of the $\sqrt[3]{5}$ (I will use $\alpha=\sqrt[3]{5}$ for the rest of this answer).
If we square $\sqrt{2}+\alpha$, and throw away any integers, we are left with $2\sqrt{2}\alpha+\alpha^2$.  Squaring this yields 
$$ 8\alpha^2+20\sqrt{2}+5\alpha$$
and we can subtract from that $20(\sqrt{2}+\alpha)$ to be left with
$$ 8\alpha^2-15\alpha $$
Squaring one more time yields (after removing the rational middle term)
$$ 320\alpha + 225\alpha^2$$
and a linear combination of the previous two expressions easily yields (after a possible rational division) $\alpha$.
Thus we have shown that $\sqrt[3]{5}=\alpha\in\mathbb{Q}(\sqrt{2}+\alpha)$, and so of course so does $\sqrt{2} = (\sqrt{2}+\alpha)-\alpha$.  In other words, $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subset\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$.
