 Solution

The irreducible cubic is separable, and its Galois group acts transitively on its three roots. The only transitive subgroups of $S_3$ areFor the cycle $\sigma :(\alpha ,\beta ,\gamma )\mapsto (\beta ,\gamma ,\alpha )$, the Lagrange resolvent for a cubic obeys $\sigma (x)={\zeta }^{2}x$. A transposition interchanging $\beta ,\gamma$ sends $x$ to $y=\alpha +\zeta \gamma +{\zeta }^2\beta$. Since $\zeta \in K$, these scalars are fixed by every automorphism. Thus the two possibilities for the image set areThese are descriptions of sets, so coincident values or a zero resolvent are not being asserted distinct.