There are many ways to do it. Which method have you learned so far?

You can use reduction modulo , or you can use the method where you find cubic resolvent and then determine the Galois group. Have you learned this?

Anyways, It is easily seen that the polynomial is irreducible over . You can try to show this as following.

It is easily seen by Rational root theorem that the polynomial has no rational root, furthermore assume that the polynomial can be written as

and then try to get a contradiction. Therefore you can conlude that the polynomial is irreducible over .

By reducing the polynomial modulo we have and modulo 5 we have

. Since The Galois group contains a trasoposition and a 3-cycle we conclude that the Galois group is