= Solution
The <Fundamental theorem of Galois theory> gives an inclusion-reversing bijection between <subgroups> $H\leq\operatorname{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto\operatorname{Gal}(L/M)$, for a finite <Galois extension>. One has $[L:L^H]=|H|$ and $[L^H:K]=[\operatorname{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.
Let $\alpha=\sqrt[3]2$ and $\zeta=e^{2\pi i/3}$. The <splitting field> is $L=\mathbb Q(\alpha,\zeta)$. A <splitting field> in <characteristic zero> is normal and separable, hence Galois. <Eisenstein criterion> at $2$ gives $[\mathbb Q(\alpha):\mathbb Q]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic <polynomial>. Thus $[L:\mathbb Q]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The <subgroup> fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition <subgroups>, so \b[it is not normal].
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