The roots of $f = x^3 -2$ are $\{2^{1/3}, a, a^2\}$, where $a = \frac{-1+\sqrt{3}i}{2}$. So let $E$ be the splitting field of $f$ over $\mathbb{Q}$, then $E = \mathbb{Q}(2^{1/3}, a)$.
Now I attempt to find $[E:\mathbb{Q}] = n$. So $[E:\mathbb{Q}] = [\mathbb{Q}(2^{1/3},a):\mathbb{Q}(2^{1/3})][\mathbb{Q}(2^{1/3}):\mathbb{Q}]$, and this is where I am a bit confused, or uncertain rather. So $f$ has an irreducible factor of degree $2$ in $\mathbb{Q}(2^{1/3})$, call it $g$, which has complex roots. This $g$ is the minimal polynomial of $a$ over $\mathbb{Q}(2^{1/3})$, so then $[\mathbb{Q}(2^{1/3},a):\mathbb{Q}(2^{1/3})] = 2$.
Now onto $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]$. The basis of $\mathbb{Q}(2^{1/3})$ over $\mathbb{Q}$ is $\{1, 2^{1/3}\}$, so then $[\mathbb{Q}(2^{1/3}):\mathbb{Q}] = 2$.
Together this gives $[E:\mathbb{Q}] = 4$.
Is this wrong? I find Galois theory quite difficult, so please be gentle.
Best Answer
Some ideas: in spite of being wrong about what the roots fo that cubic are over the rationals, it is true $\;\Bbb Q(\sqrt[3]2,\,a)\;$ is the splitting field of $\;f(x)=x^3-2\in\Bbb Q[x]\;$ over $\;\Bbb Q\;$, as it is the minimal field extension of $\;\Bbb Q\;$ containing all its roots.
Now, since $\;x^3-1=(x-1)(x^2+x+1)\;$ , we have that the minimal polynomial of $\;a\;$ over the rationals is $\;x^2+x+1\;$ ...but this polynomial remains irreducible in $\;\Bbb Q(\sqrt[3]2)[x]\;$ since $\;\Bbb Q(\sqrt[3]2)\subset\Bbb R\;$ , whereas $\;a\in\Bbb C\setminus\Bbb R\;$, and from here $\;[\Bbb Q(\sqrt[3]2,\,a):\Bbb Q(\sqrt[3]2)]=2\;$ , so altogether:
$$[\Bbb Q(\sqrt[3]2,\,a):\Bbb Q]=[\Bbb Q(\sqrt[3]2,\,a):\Bbb Q(\sqrt[3]2)][\Bbb Q(\sqrt[3]2):\Bbb Q]=2\cdot3=6$$
Observe that $\;[\Bbb Q(\sqrt[3]2):\Bbb Q]=3\;$ since $\;f(x)\;$ is irreducible over the rationals.