$\mathbf{Q}[\sqrt 5+\sqrt[3] 2]=\mathbf{Q}[\sqrt 5,\sqrt[3] 2]$? Is there a general or elegant way to approach this problem? One can show that $\sqrt 5+\sqrt[3] 2$ is a root of the hexic $x^6-6x^4-10x^3+12x^2-60x+17$, which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in $\mathbf{F}_4$, and to show it's irreducible over $\mathbf{F}_3$ already seems to require a lot of grunt work.
One thing I can do is rule out the possibility that $[\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2$, because $[\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6$ and $\mathbf{Q}(\sqrt 5,\sqrt[3] 2)$ is the compositum of $\mathbf{Q}[\sqrt 5+\sqrt[3] 2]$ and $\mathbf{Q}[\sqrt 5]$, where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out $3$, because I know of no general lower bound on the degree of a compositum. 
Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated.

It's not the most general, but I actually find the explicit approach to be relatively elegant for this problem: 
Let $\alpha=\sqrt{5}+\sqrt[3]{2}$.  Then $\alpha-\sqrt{5}=\sqrt[3]{2}$, so $(\alpha-\sqrt{5})^3\equiv \alpha^3-3\sqrt{5}\alpha^2+15\alpha-5\sqrt{5}=2$.  In other words, $(3\alpha^2+5)\sqrt{5}=\alpha^3+15\alpha-2$ or $\sqrt{5}=\dfrac{\alpha^3+15\alpha-2}{3\alpha^2+5}$.
An analagous calculation should show that the result does generalize to the case $\mathbb{Q}[\sqrt{p}+\sqrt[n]{q}]$ for any relatively prime $p,q$ and any $n\geq 2$; the fact that any powers of the quadratic are either rational or a rational multiple of the quadratic itself make it straightforward to construct this sort of explicit expression.
