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!
 A: Since $\mathbb{Q}(\sqrt{2} + \sqrt[3]{5}) \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$, it suffices to show that $[\mathbb{Q}(\sqrt{2} + \sqrt[3]{5}) : \mathbb{Q}] = 6$. The Galois closure of $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ is $N = \mathbb{Q}(\sqrt{2},\sqrt[3]{5},i)$. There are $12$ elements of the Galois group $\textrm{Gal}(N/\mathbb{Q})$. We will show that the subgroup $H$ which fixes $\mathbb{Q}(\sqrt{2} + \sqrt[3]{5})$ has order $2$, hence $[\mathbb{Q}(\sqrt{2} + \sqrt[3]{5}) : \mathbb{Q}] = 6$.
Let $\sqrt{2} + \sqrt[3]{5} = \alpha$ and let $\sigma \in \textrm{Gal}(N/\mathbb{Q})$ such that $\sigma$ fixes $\alpha$. Now the conjugate of $\sqrt{2}$ is real and the two conjugates of $\sqrt[3]{5}$ are both nonreal. It follows that $\sigma$ must fix $\sqrt[3]{5}$. Then $\sigma$ also fixes $\sqrt{2} = \alpha - \sqrt[3]{5}$. The only possibilities are thus $\sigma$ is the identity or $\sigma$ fixes $\sqrt{2},\sqrt[3]{5}$ and $\sigma(i) = -i$, hence $\vert H \vert = 2$.
A: Let $x \in \mathbb{Q}(\sqrt{2}, \sqrt[3]{5})= \mathbb{Q}(\sqrt{2})(\sqrt[3]{5})\Rightarrow x = y+z\sqrt[3]{5}$, and $y, z \in \mathbb{Q}(\sqrt{2})\Rightarrow y = a+b\sqrt{2}, z = c+ d\sqrt{2}$, and $a, b,c,d \in \mathbb{Q}$. Can you take it from here? 
A: You can also calculate $[\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}):\mathbb{Q}]$ and 
$[\mathbb{Q}(\sqrt{2},\sqrt[3]{5}):\mathbb{Q}]$. If they have the same index over $\mathbb{Q}$ and one is contained in the other they must be equal.
For the sectond part you can use a similar argument I think. 
A: Hint: $[Q(\sqrt2,\sqrt[3]5):Q]=6$, $[Q(\sqrt2,\sqrt[3]5):Q]=[Q(\sqrt2,\sqrt[3]5):Q(\sqrt2+\sqrt[3]5)][Q(\sqrt2+\sqrt[3]5):Q]$. This implies that $[Q(\sqrt2+\sqrt[3]5):Q]=1,2,3$ or 6.
$[Q(\sqrt2+\sqrt[3]5):Q]$ can't be $1$ since $u=\sqrt2+\sqrt[3]5$ is not in $Q$ it can't be 2 since for every $q\in Q$, $(u-q)^2$ is not in $Q$,
Remark that if a number satisfies $X^2+2bX+c=0, b,c\in Q$ you can write $(X+b)^2-b^2+c=0$.
it can't be 3 since $[Q(\sqrt2,\sqrt[3]5):Q(\sqrt2+\sqrt[3]5)]=2$, since in this case if $\sqrt[3]5$ is  an element of $Q(\sqrt2+\sqrt[3]5)$ so is $\sqrt2$ and $Q(\sqrt2,\sqrt[3]5)=Q(\sqrt2+\sqrt[3]5)$ which contradicts the fact that $[Q(\sqrt2,\sqrt[3]5):Q(\sqrt2+\sqrt[3]5)]=2$, thus $\sqrt[3]5$ is not an element of $Q(\sqrt2+\sqrt[3]5)$ and it is the zero of $X^3-5$, the minimal polynomial $P$ of $\sqrt[3]5$ in $Q(\sqrt2+\sqrt[3]5)$ divides $X^3-5$. This is impossible since the roots of $P$ should be $\sqrt[3]5$ and a complex root of $X^3-5$, thus $P$ should have a complex coefficient.
A: 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, $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subset\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$.
