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.

We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of
$$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$
$$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$
$$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$
simplify giving the coefficients of $x^3+x^2-6x-7$. That depends on a Kummer sum, the cubic equivalent of a Gaussian period.
For short: the Galois group of the minimal polynomial of $\xi$ is $\mathbb{Z}/(18\mathbb{Z})$, and the polynomial $x^3+x^2-6x-7$ encodes a subgroup of such a group. Notice that $\{1,7,8,11,12,18\}$ are exactly the cubic residues in $\mathbb{Z}/(19\mathbb{Z})^*$, that is generated by $2$, and $\!\!\pmod{19}$
$$2\cdot\{1,7,8,11,12,18\}=\{2,3,5,14,16,17\},$$
$$2^2\cdot\{1,7,8,11,12,18\}=\{4,6,9,10,13,15\},$$
giving the exponents associated with $c,a,b$.

We may perform the same trick with the prime $p=13\equiv 1\pmod{3}$. $\mathbb{Z}/(p\mathbb{Z})^*$ is generated by $2$ and the cubic residues $\!\!\pmod{p}$ are $\{1,5,8,12\}$, hence if we take $\xi=\exp\left(\frac{2\pi i}{13}\right)$ we have that:
$$ x_1 = \xi^1+\xi^5+\xi^8+\xi^{12} = 2\cos\left(\frac{2\pi}{13}\right)+2\cos\left(\frac{10\pi}{13}\right),$$
$$ x_2 = \xi^2+\xi^3+\xi^{10}+\xi^{11} = 2\cos\left(\frac{4\pi}{13}\right)+2\cos\left(\frac{6\pi}{13}\right),$$
$$ x_3 = \xi^4+\xi^6+\xi^7+\xi^{9} = 2\cos\left(\frac{8\pi}{13}\right)+2\cos\left(\frac{12\pi}{13}\right),$$
are conjugated algebraic numbers, roots of the polynomial
$$ p(x) = x^3+x^2-4x+1.$$
