Splitting field and field extension $\mathbb Q(j,\sqrt[3]2)$ I have to clarify some stuffs which is not clear in my mind.
I denote $j=e^{\frac{2i\pi}{3}}$.
1) $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is a field extension of degree $3$ since $X^3-2$ is the minimal polynomial of $\sqrt[3]2$. The extension is separable (since $X^3-2$ is separable) but not normal (since $j\sqrt[3]2$ is a root conjugate that it's not in $\mathbb Q(\sqrt[3]2)$). 
2) $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is the splitting field of $X^3-2$ ? And thus $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is a Galois extension since it's the splitting field of a separable polynomial.
3) $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is an extension of degree $6$ since $[\mathbb Q(\sqrt[3]2):\mathbb Q]=3$ and since $j\notin \mathbb Q(\sqrt[3]2)$ and that it's minimal polynomial on $\mathbb Q(\sqrt[3]2)$ is $X^2+X+1$, we get $[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]=2$ and thus $$[\mathbb Q(\sqrt[3]2,j):\mathbb Q]=[\mathbb Q(\sqrt[3]2,j):\mathbb Q(\sqrt[3]2)]\cdot [\mathbb Q(\sqrt[3]2):\mathbb Q]=2\cdot 3=6.$$
My questions are : What do you think about 2) ? And for 3) ?
 A: To expand on what I said in the comments:
Your argument for $3$ (and also your argument in the comments) works in this particular case, because the degree of $X^2 + X+1$ is small.
Your claim that the minimal polynomial of $j$ over $\mathbb Q(\sqrt[3]2)$ is the same as the minimal polynomial of $j$ over $\mathbb Q$ because $j\notin \mathbb Q$ works in this case, but not in general. For example, $e^{2i\pi/8}\notin \mathbb Q(i)$, but its minimal polynomial over $\mathbb Q(i)$ is $X^2-i$, not $X^4+1$, its minimal polynomial over $\mathbb Q$.

So why does this argument work in this case? I'll write out a correct argument along your lines and highlight where we avoid the issue mentioned above.
Since $j$ is a root of $f(X)=X^2+X+1$, and $f$ has coefficients in $\mathbb Q(\sqrt[3]2)$, we know that $$[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]\color{red}{\le}2.$$
Now the fact that $j\notin \mathbb Q(\sqrt[3]2)$ tells us that $$[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]\ge 2$$and the rest of your argument now applies.

If we try this same argument for the example above, the fact that $e^{2\pi/8}$ is a root of $X^4+1$ tells us that $$[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]\le 4$$and the fact that $e^{2\pi i/8}\notin \mathbb Q(i)$tells us that 
$$[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]\ge2.$$These two facts aren't enough to deduce the value of $[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]$.
