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**Boysilver** Let $\displaystyle f(x)=x^4+6x^2+3x-3$ and let $\displaystyle L$ be a splitting field of $\displaystyle f$ over $\displaystyle \mathbb{Q}$.

I can show that 73 has a cube root in $\displaystyle L$. Now I'm asked to "deduce" that the Galois group $\displaystyle G$ of $\displaystyle f$ over $\displaystyle \mathbb{Q}$ has a normal subgroup $\displaystyle H$ such that the quotient group $\displaystyle G/H$ is isomorphic to $\displaystyle S_3$, and then deduce that $\displaystyle G$ is isomorphic to $\displaystyle S_4$.

My first thought was that we can show $\displaystyle G$ is isomorphic to $\displaystyle S_4$ easily enough anyway - it's not difficult to narrow down the possibilities to either $\displaystyle A_4$ or $\displaystyle S_4$, then we can use the discriminant to rule out the first case. Once we know $\displaystyle G$ is isomorphic to $\displaystyle S_4$ then we can show the "normal subgroup" part.

But this is obviously not what the question is getting at...the idea is to presumably use the fact that 73 has a cube root. Can anyone help out? Thanks.