Whenever a polynomial is solvable by radicals, the Galois group of its splitting field must be a solvable group.
A group $G$ is solvable if there are subgroups $H_0, H_1, \dots , H_n$ such
that
$$\ 1 = H_0 ≤ H_1 ≤ H_2 ≤ \cdots ≤ H_n = G$$
with the property that each $H_i$ is normal in $H_{i+1,}$ and $H_{i+1}/H_i$ is abelian.
For $x^3-2$ $K=\mathbb Q(\zeta,2^{1/3})$ is the splitting field of the polynomial.
Now in the Galois groups lattice copied from a series of lectures on Galois theory at this point:
and it is clear that for $\mathbb Q(\zeta)$ (branch in light blue on the right diagram) the conditions of solvability are met, being a normal subgroup isomorphic to $\mathbb Z_3:$
But as it is indicated on the image, the other subgroups $\{e,t\},$ $\{e,tr\}$ and $\{e,tr^2\}$ are not normal, but are necessary to match the adjoining of $2^{1/3}$ to the base field.
So how is this lattice indicative of a solvable Galois group, when some subgroups are neither abelian, nor normal? How is the condition of solvability still applicable to this Gal group?
After the answer and comments by Jyrki Lahtonen
$\mathbb Q(ζ_3)$ is normal. And that field is not really about $x^3−2.$ That field is a splitting field of $x^2+x+1$ over $\mathbb Q.$ Here it is just a stepping stone along the path to the splitting field of $x^3−2.$ Whenever you have a tower of finite extensions (in characteristic zero) $K⊂F⊂L$ with $F/K$ and $L/K$ splitting fields of some polynomials, then $Gal(L/F)$ is automatically a normal subgroup of $Gal(L/K).$ Another part of the picture is that the extension $\mathbb Q(ζ,2^{1/3})/\mathbb Q(ζ)$ is also normal. And its Galois group is cyclic of order three, generated by the automorphism $r$ in your other question.
Best Answer
The criterion of solvability doesn't say that any chain of subgroups must fulfill these requirements of normality and abelianness. Just that there is at least one chain that does. You have found such a chain (the light blue branch), so $S_3$ is solvable. The fact that other chains of subgroups fail to meet the solvability criterion does not change that in the slightest.