Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation 
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is 
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got 
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I tried to factorise it, got this 
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)$
$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)$
 Should I now solve cubic?
 A: $\pm 2,\pm 3,\pm 5$ are precisely twice the cubic residues $\!\pmod{19}$. Let $\omega=\exp\left(\frac{2\pi i}{19}\right)$. $\omega$ is al algebraic number over $\mathbb{Q}$ with degree $18$, hence by Galois theory both
$$ \sum_{k\in\{\pm 1,\pm 7,\pm 8\}}\omega^k,\qquad \sum_{k\in\{\pm 2,\pm 3,\pm 5\}}\omega^k $$
(related to the Kummer sums $\!\!\pmod{19}$) are algebraic numbers over $\mathbb{Q}$ with degree $3$. Indeed, the minimal polynomial of the second sum
is $z^3+z^2-6z-7$, and the second sum is exactly $2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right)$.
A: I created this equation by the following way.
$$3^n\equiv1,3,9,8,5,15,7,2,6,...\!\pmod{19}$$
Now, let
$$x_1=2\left(\cos\frac{2\cdot1\pi}{19}+\cos\frac{2\cdot8\pi}{19}+\cos\frac{2\cdot7\pi}{19}\right),$$
$$x_2=2\left(\cos\frac{2\cdot3\pi}{19}+\cos\frac{2\cdot5\pi}{19}+\cos\frac{2\cdot2\pi}{19}\right)$$ and 
$$x_3=2\left(\cos\frac{2\cdot9\pi}{19}+\cos\frac{2\cdot15\pi}{19}+\cos\frac{2\cdot6\pi}{19}\right).$$
Thus, $x_1$, $x_2$ and $x_3$ they are roots of the cubic equation with rational coefficients.
Easy to  show that it's $$x^3+x^2-6x-7=0.$$
Also, easy to show that $$-x_1=\sqrt{4-x_2},$$
$$x_2=\sqrt{4-x_3}$$ and $$-x_3=\sqrt{4-x_1}$$
which gives an equation with radicals.
I posted this equation here: https://artofproblemsolving.com/community/c6h69475
