resolvent cubic of . The resolvent cubic of p(x) is
. Since , we see that f(x) factors as linear times irreducible quartic. Check p(x) irreducible over . Thus (link).
If you see the structure, you have a cyclic group is normal in . A little bit of comutation shows that the splitting field of p(x) over Q is and K is a galois extension of F, i.e. , Gal(K/F)=[K:F] (It is because a cyclic extension is a galois extension).