galois group of cubic polynomial with 3 real roots--no discriminant

I know it's a duplicate, but the other one is a year old and got an answer using methods far beyond the typical first year abstract algebra.

If I have an irreducible polynomial over $\mathbb{Q}$ with 3 real roots, is there a way to calculate its Galois group without using the discriminant? The only way I can think of is adjoining one root $\alpha$ and trying to see if that's the splitting field over $\mathbb{Q}$ just by using algebraic manipulations, but that is a very non reliable way.

Any suggestions? The polynomial I have in mind is $x^3-3x+1$.

A way to avoid the discriminant is to notice that we are dealing with the minimal polynomial of $2\cos\frac{2\pi}{9}$: since the third Chebyshev polynomial of the first kind is $T_3(x)=4x^3-3x$, given $p(x)=x^3-3x+1$ we have:
$$p(2\cos\theta) = 2 T_3(\cos\theta)+1 = 2\cos(3\theta)+1$$ so the roots of $p(x)$ are $\zeta_1=2\cos\frac{2\pi}{9},\zeta_2=2\cos\frac{4\pi}{9},\zeta_3=2\cos\frac{8\pi}{9}$ (they are distint numbers since $\cos x$ is a decreasing function on $(0,\pi)$). The cosine duplication formula hence gives that the Galois group of $p$ is a cyclic group of order $3$, since: $$\varphi(x): x \to 2\cdot\left(2\left(\frac{x}{2}\right)^2 - 1\right)=x^2-2$$ brings $\zeta_1$ to $\zeta_2$, $\zeta_2$ to $\zeta_3$ and $\zeta_3$ to $\zeta_1$, so $\mathbb{Q}(\cos(2\pi/9))$ is the splitting field of $p$.
• Nice, but this is special to the particular polynomial $x^3-3x+1$. General method? – Lubin Aug 8 '15 at 3:24
• Then one may just try to apply the same approach and write the roots as functions of the cosine of something, but that something is a rational multiple of $\pi$ just in a few cases. The discriminant is a powerful tool, so I don't even know if it is possible toavoid its computation, in the general case. – Jack D'Aurizio Aug 8 '15 at 16:56