<s>[INST] The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$ I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$. 
I managed to find that it is a root of the polynomial $x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17$. This is where I run in to some problems. I don't know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?
 [/INST] It's enough to show that $x^3-5$ is irreducible over $\mathbb{Q}(\sqrt{2})$. If it weren't, then it would have a root in $\mathbb{Q}(\sqrt{2})$.  Since it has a unique real root, we must have $\sqrt[3]{5}\in \mathbb{Q}(\sqrt{2})$. So $(a+b\sqrt{2})^3=5$, and therefore $(a-b\sqrt{2})^3=5$, hence $a+b\sqrt{2}=a-b\sqrt{2}=\sqrt[3]{5}$, and so $\sqrt[3]{5}\in \mathbb{Q}$, contradiction.
$\bf{Added:}$  I am uncomfortable with my answer, since it assumes $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{5}) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5})$, which is the crux of the problem. If that is proved, then the degree has to be divisible by $2$, and $3$, so it is necessarily $6$. So let's just show that $\alpha = \sqrt{2}+ \sqrt[3]{5}$ generates both roots. Now, we have the equality $(\alpha - \sqrt{2})^3= 5$ from which we can obtain $\sqrt{2}$ as a rational fraction in $\alpha$. In detail:
$$\alpha^3 - 3 \alpha^2 \sqrt{2} +3 \alpha \cdot 2 - 2 \sqrt{2} = 5$$
or
$$\frac{\alpha^3 +6 \alpha - 5 }{3 \alpha^2 + 2} = \sqrt{2}$$
By the way, the equation that we get for $\alpha$
$$(x^3 + 6 x -5)^2 - 2(3x^2 + 2)^2=0$$
is the equation in the posting.
</s>