 Solution

Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent isThe rational root theorem shows it is irreducible over $\mathbb{Q}$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently