Roots of $y=x^3+x^2-6x-7$ I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?
Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+2\cos\left(\frac {16\pi}{19}\right)$ and $2\cos\left(\frac {8\pi}{19}\right)+2\cos\left(\frac {12\pi}{19}\right)+2\cos\left(\frac {18\pi}{19}\right)$.
Anything helps. I don't think substituting $x$ with something like $t+t^{-1}$ will work.

By the translation $x\to x-\frac13$, you cancel the square term.
$$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$
Then multiplying by $27$ and replacing $x\to3x$, you get
$$x^3-57x-133=0.$$
Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\theta)$ appear),
$$152\sqrt{19}\cos^3(\theta)-114\sqrt{19}\cos(\theta)-133=38\sqrt{19}(4\cos^3(\theta)-3\cos(\theta))-133=0$$
so that
$$\cos(3\theta)=\frac{133}{38\sqrt{19}}.$$
Retrieving the three roots from here is straightforward.
This solution is due to François Viète. You can see it as an instance of the famous "angle trisection" problem.

Another approach is by decomposing $x=u+v$. You can develop the equation as
$$u^3+3uv(u+v)+v^3-57(u+v)-133=0.$$
Then setting $3uv=57$, two terms cancel out and you end up with the system
$$\begin{cases}uv=19\iff u^3v^3=6859\\u^3+v^3=133.\end{cases}$$
This is a sum/product problem and
$$u^3,v^3=\frac{133\pm\sqrt{133^2-4\cdot6850}}2.$$
But the story ends here, because these solutions are complex numbers, which is harmless in itself, but you need to take their cubic roots and this requires the use of... trigonometric functions. This is called the casus irreductibilis as radicals don't suffice.
