$\mathbf{Q}[\sqrt 5+\sqrt[3] 2]=\mathbf{Q}[\sqrt 5,\sqrt[3] 2]$? 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.

Apologies if this duplicates content in the nice answer of orangeskid. Here is another approach, also using Galois theory. 
Let $K$ be the splitting field of the polynomial $(x^{2}-5)(x^{3}-2)$ over $\mathbb{Q}$, and let $\alpha = \sqrt[3]{2}+\sqrt{5}$. One can see that $K = \mathbb{Q}(\sqrt{5}, \sqrt[3]{2}, \xi)$, where $\xi$ is a primitive third root of unity, and hence is the normal closure of the subfield $L := \mathbb{Q}(\sqrt{5}, \sqrt[3]{2})$. Now, $[L:\mathbb{Q}] = 6$, since $\deg(\sqrt{5}) = 2$ and $\deg(\sqrt[3]{2}) = 3$ both divide $[L:\mathbb{Q}]$, and $\mathbb{Q}(\sqrt[3]{2}+\sqrt{5})$ is clearly a subfield of $L$. If we can show that $\deg(\alpha) = 6$, the conclusion follows.
Since $L \subset \mathbb{R}$ and $\xi \notin \mathbb{R}$, it follows that $[K:L] = 2$, whence $[K:\mathbb{Q}] = 12$. Then $G := \mathrm{Gal}(K/\mathbb{Q})$ has $12$ elements, and any element $\sigma \in G$ is determined by what it does to the generators $\sqrt{5}, \sqrt[3]{2}, \xi$ of $K$ over $\mathbb{Q}$. Any $\sigma \in G$ must satisfy $\sigma(\sqrt{5}) \in \{\pm \sqrt{5}\}, \sigma(\sqrt[3]{2}) \in \{\xi^{i}\sqrt[3]{2}\}_{i = 0}^{2}, \sigma(\xi) \in \{\xi, \xi^{2}\}$, so counting, there are at most 12 possible automorphisms, arising from these possible assignments on generators. Since we have already determined $|G| = 12$, every assignment on generators described above must give rise to an automorphism of $K$ over $\mathbb{Q}$. Hence, let $\sigma, \tau \in G$ be the automorphisms defined on generators by 
$$\sigma(\sqrt[3]{2}) = \xi\sqrt[3]{2}, \sigma(\sqrt{5}) = \sqrt{5}, \sigma(\xi) = \xi$$
$$\tau(\sqrt[3]{2}) = \sqrt[3]{2}, \tau(\sqrt{5}) = -\sqrt{5}, \tau(\xi) = \xi$$
Then the Galois orbit $G\alpha$ contains $\{\pm \sqrt{5} + \xi^{i}\sqrt[3]{2} \mid i = 0, 1, 2\}$, so $|G\alpha| \geqslant 6$. Since $|G\alpha| \leqslant 6$, we obtain the equality $|G\alpha| = 6$. Further, as the minimal polynomial of $\alpha$ over $\mathbb{Q}$ factors as $\prod_{s \in G\alpha} (X-s)$, it follows that $\deg(\alpha) = 6$, as desired. 
