Finding all reals such that two field extensions are equal. 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). 
 A: The field has degree 6 over the rationals. Any element $w$ of degree 6 will generate the field. 
Now, every element of the field has degree 1, 2, 3, or 6. The only elements of degree 1 are the rationals. The only elements of degree 2 are those of the form $a+b\sqrt2$ (although it takes some work to check this). The only elements of degree 3 are those of the form $a+b\root3\of5+c\root3\of{25}$ (again, this takes some checking). It follows that the generators are all the elements $a+b\sqrt2+c\root3\of5+d\sqrt2\root3\of5+e\root3\of{25}+f\sqrt2\root3\of{25}$ except those with $b=c=d=e=f=0$, those with $c=d=e=f=0$, and those with $b=d=f=0$. 
A: If we take $u = \sqrt{2} + \sqrt[3]{5}$, such a $u$ almost always turns out to work.  In fact let's try if a rational linear combination of $\sqrt{2}$ and $\sqrt[3]{5}$ will work. Let us now write $u$ as $u = a\sqrt{2} + b\sqrt[3]{5}$ for rationals $a$ and $b$.
Clearly we have that $\Bbb{Q}(u)\subseteq \Bbb{Q}(\sqrt{2},\sqrt[3]{5})$. To show the other inclusion, we just need to show that say $\sqrt{2} \in \Bbb{Q}(u)$ for then $\sqrt[3]{5} = \frac{a\sqrt{2} + b\sqrt[3]{5} - a\sqrt{2}}{b}$  will be in $\Bbb{Q}(u)$. Here is a quick and easy way of doing this:
Write $u = a\sqrt{2} + b\sqrt[3]{5}$ so that $(\frac{u - a\sqrt{2}}{b})^3  = 5$. We can assume that $a$ and $b$ are simultaneously not zero for then the proof becomes redundant. Then expanding the left hand side by the binomial theorem we get that
$$ u^3  - 3\sqrt{2}u^2a + 6ua^2  - 2a^3\sqrt{2} = 5.$$
Rearranging, we get that 
$$\sqrt{2} = \frac{u^3 + 6ua^2 -5}{    3u^2a + 2a^3     }.$$
Since $\Bbb{Q}(u)$ is a field the right hand side is in $\Bbb{Q}(u)$ so that $\sqrt{2}$ is in here. Done!
