After reading this question Splitting field of $x^3-5 \in \mathbb{Q}[X]$. Galois group and fields?, I tried to think about the following question: Find the splitting field $L$ of the polynomial $X^3-10\in \mathbb{Q}[X]$. And I try to figure out if the Galois extension $L$ over $\mathbb{Q}$ is abelian.
My idea is that
(1) splitting field $L$:
Note that all roots are $10^{1/3}\omega^i$ for $i=0,1,2$ where $\omega^3=1$. So the splitting field is $L=\mathbb{Q}(10^{1/3}, \omega)$.
(2) order of extension $[L:\mathbb{Q}]$:
since $\;x^3-1=(x-1)(x^2+x+1)\;$ , we have that the minimal polynomial of $\;\omega\;$ over the rationals is $\;x^2+x+1\;$
This polynomial remains irreducible in $\;\Bbb Q(\sqrt[3]10)[x]\;$ since $\;\Bbb Q(\sqrt[3]10)\subset\Bbb R\;$ , whereas $\;\omega\in\Bbb C\setminus\Bbb R\;$, and from here $\;[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q(\sqrt[3]10)]=2\;$ , [Question: Am I right in saying this?]
Also, $\{1, 10^{1/3}, 10^{2/3}\}$ is a basis for $\Bbb{Q}(10^{1/3})$ over $\Bbb{Q}$, so $[\Bbb{Q}(10^{1/3}):\Bbb{Q}]=3$.
So altogether:
$$[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q]=[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q(\sqrt[3]10)][\Bbb Q(\sqrt[3]10):\Bbb Q]=2\cdot3=6$$
(3) Because this is a splitting field over $\mathbb{Q}$, the extension is normal and separable, hence Galois, so the Galois group has order 6. The splitting field of a degree $n$ polynomial is a subgroup of $S_n$, and $\vert S_3\vert=6$, so has to be it. So $$ Gal(L/\mathbb{Q})\cong S_3 $$ which is not abelian. [Question: Am I right in saying this?]
What you did is fine, except the miscalculation of the minimal polynomial of $\omega$, as pointed out in the comment by Gerry Myerson.
There is a more general result along the line: [Hungerford, Chapter V, Theorem 4.12]
The proof is very simple: Let $\alpha$ be (any) root of $f$, then $[\mathbb Q[\alpha]:\mathbb Q]=p$, and by Galois correspondence, the Galois group $G$ contains a subgroup of index $p$, hence $p\mid |G|$, and by Cauchy's theorem, there exists an element of order $p$ in $G$. As $G$ can be regarded as a subgroup of $S_p$, and in $S_p$, an element of order $p$ must be a $p$-cycle. The complex conjugate will keep all the real roots fixed and interchange the two imaginary roots, therefore is a transposition in $S_p$. Now it's a group theory exercise to show that a $p$-cycle and a transposition generate $S_p$ (In the special case of $p=3$, this is easy by both $2$ and $3$ divide $|G|$, as you have done).
In particular, for any polynomial $x^3-n$ where $n\in\mathbb Z$ is not a perfect cubic number, the Galois group is isomorphic to $S_3$.