= Solution
The irreducible cubic is separable, and its <Galois group> acts transitively on its three roots. The only transitive subgroups of $S_3$ are
$$
\boxed{A_3\cong C_3\quad\hbox{and}\quad S_3.}
$$
For the cycle $\sigma:(\alpha,\beta,\gamma)\mapsto(\beta,\gamma,\alpha)$, the <Lagrange resolvent for a cubic> obeys $\sigma(x)=\zeta^2x$. 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 are
$$
\boxed{\{x,\zeta x,\zeta^2x\}\quad\text{for }A_3,\qquad
\{x,\zeta x,\zeta^2x,y,\zeta y,\zeta^2y\}\quad\text{for }S_3.}
$$
These are descriptions of sets, so coincident values or a zero resolvent are not being asserted distinct.
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