Solvability of x^n - a

Remarks

If $a = 0$, the polynomial $x^n$ splits over $K$ and the group is trivial. If $a \ne 0$, let $L$ be a splitting field and $\beta \in L$ a root; the roots are $\beta \zeta^k$, where $\zeta$ is a primitive $n$-th root of unity, so $\zeta \in L$. Put $E = K(\zeta)$; $E/K$ is a cyclotomic extension, and $\Gal(E/K)$ embeds in $(\Z/n\Z)^\times$, so it is abelian. The map $\sigma \mapsto \sigma(\beta)/\beta$ embeds $\Gal(L/E)$ in the cyclic group of $n$-th roots of unity, so $L/E$ is a cyclic extension. Then $\Gal(L/E)$ is a normal subgroup of $\Gal(L/K)$ with abelian quotient $\Gal(E/K)$, so $\Gal(L/K)$ is solvable. Example: for $x^3 - 2$ over $\Q$, the subgroup is cyclic of order $3$ and the quotient has order $2$, giving $S_3$.

Going deeper

For a student

This is the base case of the easy direction of Galois' theorem on solvability. The two-step tower $K \subseteq K(\zeta) \subseteq L$ is the model for every radical extension: first roots of unity, then pure roots. The same argument shows that $\Gal(L/K)$ embeds in the group of maps $x \mapsto ux + v$ on $\Z/n\Z$ with $u$ a unit.