Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$ 
Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$

My idea was to find the minimal polynomial of $\sqrt[3]{5}+\sqrt{2}$ over $\mathbb{Q}$ and to show that $\deg p(x)=6$
Attempt:
Let $u:=\sqrt[3]{5}+\sqrt{2}\\
u-\sqrt[3]{5}=\sqrt 2\\
(u-\sqrt[3]{5})^2=2\\
u^2-2\sqrt[3]{5}u+5^{2/3}-2=0\\
u^2-2-5^{2/3}=2\sqrt[3]{5}u\\
(u^2-2-5^{2/3})^3=2^3\cdot 5 \cdot u$
I'm stuck here 
Here Wolfram's result
My previous question over $\mathbb{Q}(\sqrt[3]{5})$

First, $p:=x^3-5$ and $q:=x^2-2$ are the minimal polynomials of $\sqrt[3]{5}$ and $\sqrt{2}$ over $\mathbb{Q}$ since they are monic and using Eisenstein's criterion, they are irreducible over $\mathbb{Q}$. Then, the following is an annihilator polynomial with rational coefficients of their sum:
$$\textrm{res}_y(p(y),q(x-y))=x^6 - 6 x^4 - 10 x^3 + 12 x^2 - 60 x + 17.$$
Indeed, $p(y)$ and $q(\sqrt[3]{5}+\sqrt{2}-y)$ both vanish on $\sqrt{2}$. To conclude, it suffices to see that the above polynomial is irreducible over $\mathbb{Q}$ and I have no trick to do so: reduction mod $2$ and $3$ fail, the reduced polynomial has a root. Feel free to share one.
