 Solution

If $a=0$, the splitting field is $K$ and its Galois group is trivial. Otherwise choose a root $\alpha$ and a primitive $n$th root of unity $\zeta \in K$. Every root is ${\zeta }^j\alpha$, so $L=K(\alpha )$. Each automorphism is determined by $\sigma (\alpha )/\alpha \in {\mu }_n$. This is an injective group homomorphism, since the roots of unity are fixed by every automorphism. A subgroup of the cyclic group ${\mu }_n$ is cyclic. Thus $\mathrm{Gal}(L/K)$ is cyclic, with order dividing $n$.