Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$ I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, just in case):
The problem is:  

a) Prove $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $\mathbb{Q}$ of degree 6.

Done: I know it has degree $\leq 6$ because $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ which has degree 6; then I explicitly found the polynomial by solving a 6-equation linear system, and Wolfram Alpha proved it irreducible (btw: how can I prove it by hand?). The polynomial is $t^6-6t^4-10t^3+12t^2-60t+17$.

b) What's its degree over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$?

It is this part b) which I can't solve. Of course its degree is $\leq 6$ in both cases, but I don't know what else to do.
 A: Bruno: The degree over ${\mathbb Q}(\sqrt2)$ is 3: Consider $(x-\sqrt 2)^3-5$. Since $\root 3\of 5$ is in ${\mathbb Q}(\sqrt 2)(\sqrt 2+\root 3\of 5)$, the only other option is that it already belongs to ${\mathbb Q}(\sqrt 2)$. But this is impossible, since its minimal polynomial is $x^3-5$ over ${\mathbb Q}$ and ${\mathbb Q}(\sqrt 2)$ is an extension of degree 2.
[ By the way, one can use this and the tower law, to show that ${\mathbb Q}(\sqrt 2+\root 3\of 5)$ has degree 6 over ${\mathbb Q}$. The irreducibility of your polynomial then follows for free. ]
Similarly, the degree over ${\mathbb Q}(\root 3\of 5)$ is 2.
A: For part b, we can find the minimal polynomial in both cases.
Consider $a =\sqrt{2} + \sqrt[3]{5}$ over $\mathbb{Q}(\sqrt[3]{5})$. Notice that
$ (x - \sqrt[3]{5})^2 -2 = 0$
is a degree two polynomial in $\mathbb{Q}(\sqrt[3]{5})[x]$ that $a$ is a root of. We know that $a \notin \mathbb{Q}(\sqrt[3]{5})$, so the degree of its minimal polynomial must be greater than 1, and now we have a polynomial of degree 2 that it is a root of, so its minimal polynomial must be that one above. Therefore $a$ has degree two over $\mathbb{Q}(\sqrt[3]{5})$.
Similarly, consider $a$ over $\mathbb{Q}(\sqrt{2})$. Notice that
$ (x - \sqrt{2})^3 - 5 = 0 $
is a degree 3 polynomial in $\mathbb{Q}(\sqrt{2})[x]$ that $a$ is a root of, and hence the minimal polynomial, which must divide this polynomial, can have degree either 2, or 3. But, if it has degree two, then it is attained from the above one by dividing by $x - a$. We get by polynomial division that
$ (x - \sqrt{2})^3 - 5 = (x - \sqrt{2} - \sqrt[3]{5})(-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2) $
and $-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2 \notin \mathbb{Q}(\sqrt{2})[x]$ because the $x$ coefficient is not in $\mathbb{Q}(\sqrt{2})$. If it were, then as $2\sqrt{2} \in \mathbb{Q}(\sqrt{2})$, we would get that 
$2 \sqrt{2} + (-2\sqrt{2} + \sqrt[3]{5}) = \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2})$
which is impossible. Thus that quadratic polynomial cannot be the minimal polynomial for $a$. Hence, the minimal polynomial of $a$ has degree 3.
