I have found that $\alpha = \sqrt{2} + \sqrt[3]{5}$ is a root of $f(x) = x^6-6x^4-10x^3+12x^2-60x+17$. I don't know if this is the minimum polynomial of $\sqrt{2} + \sqrt[3]{5}$ above $\mathbb{Q}$. I have found also that $[(\mathbb{Q}(\alpha))(\sqrt{2}):\mathbb{Q}(\sqrt{2})]=3$. This may it help. I have tried to prove that $f(x)$ is irreducible above $\mathbb{Q}$ or also to prove that $\sqrt{2} \in \mathbb{Q}(\alpha)$ but I haven't been able to do it. Do you know how to do it?
Generalization: The problem will be solved if $\mathbb{Q}(\alpha) = (\mathbb{Q}(\sqrt{2}))(\sqrt[3]{5})$. Is this always true for every couple of radicals?
Foot note: a question very similar to this has already been asked, but without solution, here the old thread Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$
I kind of wonder how you found this polynomial, that must have been quite some work. Let me show you an alternative way. Let $\zeta$ be a $3$rd unit root. Then we examine the Galois extension $$ \mathbb{Q}(\sqrt{2},\sqrt[3]{5},\zeta) $$ of degree $2\cdot3\cdot 2 = 12$. This is Galois, as any automorphism must send $\sqrt{2}$ to either itself or $-\sqrt{2}$ and it must permutate the roots of $X^3-\sqrt[3]{5}$, giving us $2\cdot 3! = 12$ automorphisms. Now the minimal polynomial equals the polynomial whose roots are those elements that $\sqrt{2} + \sqrt[3]{5}$ can be mapped to by elements of the Galois group. In this case we get that the minimal polynomial equals $$ (X-(\sqrt{2} + \sqrt[3]{5}))(X-(\sqrt{2} + \zeta\sqrt[3]{5}))(X-(\sqrt{2} + \zeta^2\sqrt[3]{5}))(X-(-\sqrt{2} + \sqrt[3]{5}))(X-(-\sqrt{2} + \zeta\sqrt[3]{5}))(X-(-\sqrt{2} + \zeta^2\sqrt[3]{5})). $$