Let $f(x) \in \mathbb Q[x]$ be an irreducible cubic polynomial with three real roots $\alpha, \beta, \gamma \in \mathbb R$ such that $\alpha$ is negative and $\beta$ is positive. Prove that the Galois group of the degree 6 polynomial $p(x) = f(x^2)$ is non-abelian.
Over $\mathbb C$, the polynomial $p$ splits into linear factors $p(x) = (x\pm \sqrt\alpha)(x\pm \sqrt\beta) (x\pm \sqrt\gamma), $ so $K = \mathbb Q(\sqrt \alpha, \sqrt\beta, \sqrt\gamma)$ is the splitting field of $p(x)$ with $\sqrt \alpha \in \mathbb C\setminus \mathbb R$.
Setting $G$ to be the Galois group of $K/\mathbb Q$, we see that the complex conjugate automorphism $\sigma$ mapping $\sqrt \alpha \mapsto - \sqrt\alpha$ and fixing everything else lies in $G$. Now, if I could show that there exists $\tau$ in $G$ such that $\tau(\sqrt\beta) = \sqrt\alpha$, then $\sigma\tau(\sqrt \beta) = -\sqrt\alpha \neq \sqrt \alpha = \tau \sigma(\sqrt \beta)$ would show that $G$ is not abelian. However, we can not make sure that $p$ is still irreducible over $\mathbb Q$ and thus $G$ may not be transitive on the roots. How should I proceed here?