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 $\mathbb{Q}(\sqrt[3]{5})$

There is no need to compute minimal polynomials.
Let $\alpha = \sqrt[3]{5}+\sqrt{2}$. Then $(\alpha-\sqrt{2})^3=5$ and so $\alpha^3 + 6 \alpha -5 =\sqrt{2} (3\alpha^2 + 2)$, which gives
$(\alpha^3 + 6 \alpha -5)^2 =2 (3\alpha^2 + 2)^2$.
Therefore,
$\alpha$ is a root of a polynomial of degree $6$ and so 
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right] \le 6$.
On the other hand,
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]$
is a multiple of $6$ because
$\left[\mathbb{Q}(\sqrt[3]{5}):\mathbb Q\right]=3$
and
$\left[\mathbb{Q}(\sqrt{2}):\mathbb Q\right]=2$.
Therefore,
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right] = 6$.
