So $f=x^3-3x-1\in \Bbb Q[x]$ is irreducible by rational roots test.
The notes say that let $\alpha$ be a root, and then we get the roots $\alpha^2-\alpha-2$ and $2-\alpha^2$. I feel like an idiot being unable to solve for these two, no idea why I can't get them.
So letting $\alpha$ be a root of $x^3-3x-1$ we have $\alpha^3-3\alpha-1=0$, and we can try dividing $(x-\alpha)$ out of $x^3-3x-1$:
$$\frac{x^3-3x-1}{x-\alpha} = x^2+\alpha x + (\alpha^2-3) + \left(\frac{\alpha^3-3\alpha-1}{x-\alpha} = 0\right) = 0$$
So we have $x^3-3x-1=(x-\alpha)(x^2+\alpha x+ (\alpha^2-3))$
So we want to find the two other roots in $x^2+\alpha x+ (\alpha^2-3)$, quadratic method is a mess, how should I proceed?
Since $T_3(x)=4x^3-3x$, our polynomial is just: $$ 2\cdot T_3\left(\frac{x}{2}\right)-1, $$ and its roots are given by $2\cos\left(\frac{\pi}{9}\right),2\cos\left(\frac{5\pi}{9}\right)$ and $2\cos\left(\frac{7\pi}{9}\right)$. The relations between the roots are so given by the cosine duplication formula. Have a look at Chebyshev polynomials of the first kind.