Determine the Galois group of $x^3 + 3x^2 - 1$ over $\mathbb{Q}$ So the only possible roots in $\mathbb{Q}$ are $1$ or $-1$ and neither are roots. So all I rely know is that the group is isomorphic to $S_3$ or $A_3$ and the polynomial has no rational roots.
Also this problem is very similar to this problem
Galois groups of $x^3-3x+1$ and $(x^3-2)(x^2+3)$ over $\mathbb{Q}$
So I am trying to apply the answer for this question, but to be honest I am not sure what the discriminate means?
 A: Substituting $x=y-1$ for $x$ yields the well-known equation $y^3-3y+1=0$, whose roots are $2cos(2\pi n/9)$ for $n=1, 4, 7$. This equation has Galois Group $A_3$; in fact, if a root is $\alpha$, then the other two roots are $\alpha ^2-2$ and $2-\alpha-\alpha ^2$, showing that adding $\alpha$ to the rationals also adds the other two roots.
A: Let $$P(x) = x^3 + 3x^2 - 1$$
First of all lets check that $P(x)$ is irreducible. If it factors at all it will have a linear factor. So we can show it is irreducible by showing it has no rational roots, this is done with the rational root theorem.
This is a degree 3 irreducible polynomial so the Galois group will be $S_3$ or $C_3$.
One tool that we can use to differentiate between these two Galois groups is the discriminant:
$$\operatorname{disc}(x^3 + b x^2 + c x + d) = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$$.
The Galois group will be $C_3$ if and only if the discriminant is a square.
In our case the discriminant is $81 = 9^2$ so the Galois group is $C_3$.
