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 $\mathbb{Q}(\sqrt[3]{5})$
 A: 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,
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right] = 6$.
A: Here is a method of attacking the problem that has a lot in common with what @C. Falcon has offered, and in addition shows the irreducibility of the polynomial $F(x)=x^6-6x^4-10x^3+12x^2-60x+17$.
First, look at $R=\Bbb Z[\sqrt2\,]$, the integers of $\Bbb Q(\sqrt2\,)$, and the polynomial ring $R[x]$. Since $5$ is still prime in $R$, $fx)=x^3-5$ is irreducible by Eisenstein, and is the minimal polynomial for $\sqrt[3]5$ over $R$. It follows that $g=f(x-\sqrt2)=x^3-3\sqrt2x^2+6x-2\sqrt2-5$ is the minimal polynomial for $\sqrt[3]5+\sqrt2$ in $R[x]$.
Now let $\bar g$ be the conjugate of $g$ (by applying the automorphism $\sqrt2\to-\sqrt2$ to the coefficients of $g$), and notice that
$$
g\bar g=x^6-6x^4-10x^3+12x^2-60x+17=F(x)\,.
$$
For the moment, do nothing but admire this factorization in the unique factorization domain $\Bbb Q(\sqrt2)[x]$. Finished admiring? Now notice that on the left-hand side of the equation, we have two $\Bbb Q(\sqrt2\,)$-irreducible polynomials, and on the right, our $\Bbb Q$-polynomial $F$, which is evidently satisfied by $\sqrt[3]5+\sqrt2$. This factorization is the only factorization of $F$ over $\Bbb Q(\sqrt2\,)$. Since every $\Bbb Q$-factorization of $F$ is a fortiori a $\Bbb Q(\sqrt2\,)$-factorization, there is no $\Bbb Q$-factorization of $F$.
A: Hint: $[(Q\sqrt[3]5+\sqrt2):Q]=[(Q\sqrt[3]5+\sqrt2):Q(\sqrt2][Q(\sqrt2):Q]$.
$[Q(\sqrt2:Q]=2$
and $((\sqrt[3]5+\sqrt2)-\sqrt2)^3=5$, so $\sqrt[3]5+\sqrt2$ is a root of $(X-\sqrt2)^3-5$ in $Q(\sqrt2)$. Show that $\sqrt[3]5+\sqrt2$ is not in $Q(\sqrt2)$ and deduce that $[(Q\sqrt[3]5+\sqrt2):Q(\sqrt2)]=3$.
A: 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.
