Let $f(x)=x^3-px+q$ be an irreducible polynomial in $\mathbb Q[x]$. Denote its complex zeros by $a_1, a_2, a_3$. Define $\det(f)=(a_1-a_2)^2(a_2-a_3)^2(a_3-a_1)^2$. It turns out $\det(f)=4p^3-27q^2$, which is rational.
Denote $K=\mathbb Q(a_1,a_2,a_3)$, a field extension of the rationals.
I've already proven that: $\det(f) \neq 0$, $[K:\mathbb Q]=3$ or $6$, which turns out by the theorem for towers of fields to depend on $[\mathbb Q(a_1,a_2):\mathbb Q(a_1)]$, which equals $1$ or $2$.
Question: Prove that $\sqrt{\det(f)}\notin\mathbb Q$ $\implies$ $[K:\mathbb Q]=6$.
I tried proving this, but I couldn't do it. I could do it directly or by contraposition. Could somebody help me? Thanks a lot!
Hint: It follows from the definition that $\sqrt{\mathrm{Det}(f)}=|(a_1-a_2)(a_2-a_3)(a_3-a_1)|\in K$. Now use the fact that $\mathrm{Det}(f)\in\mathbb Q$.