Roots of $y=x^3+x^2-6x-7$

Question

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.

Answer 1

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 $c,a,b$.

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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 $$ p(x) = x^3+x^2-4x+1.$$

Answer 2

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.

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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.