$x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ Prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a solution of the equation
$$4x^{3}+ 2x^{2}- 7x- 5= 0$$
My try:
If $x_{0}$, $x_{1}$, $x_{2}$ be the solutions of the equation then
$$\left\{\begin{matrix}
x_{0}+ x_{1}+ x_{2} = -\frac{1}{2}   \\ 
x_{0}x_{1}+ x_{1}x_{2}+ x_{2}x_{0}= -\frac{7}{4}  \\ 
x_{0}x_{1}x_{2} = -\frac{5}{4}
\end{matrix}\right.$$
I can' t continue! Help me!

As above $ 4x^{3}+ 2x^{2}- 7x- 5$ has a rational root of -1 so the problem reduces after polynomial division to proving $\cos(\frac{2 \pi}{21}) + \cos(\frac{8 \pi}{21}) + \cos(\frac{10 \pi}{21})$ is a root of  $4x^{2} - 2x - 5$ 
First some observations since these are all 21st roots of unity
Let $\omega = \frac{2 \pi}{21}$ the first root.
Then since the sum of all the roots is 0, the sum of all the cosines of the roots must also be 0. Also in the odd roots of unity half the of roots are symmetrical to the other ones across the x axis.   And from this 3 things fall out:
From the thirds:  $cos(7 \omega) = -\frac{1}{2}$
From the sevenths: $cos(3 \omega) + cos(6 \omega) + cos(9 \omega) = -\frac{1}{2}$
And left over: $cos(\omega) + cos(2 \omega) + cos(4 \omega) + cos(5\omega) + cos(8 \omega) + cos(10 \omega) = \frac{1}{2}$
(Note: using the vieta formula $x_0 + x_1 = \frac{1}{2}$ the 2nd root must be $cos(2 \omega) +cos(8 \omega) + cos(10 \omega)$ from this plus above.) 
Now you just have to plug $cos( \omega) + cos(4 \omega) + cos(5 \omega)$ into the polynomial $4x^2 - 2x$ and see what simplifies.
To start with $4x^2$ term:
$4[cos( \omega) + cos(4 \omega) + cos(5 \omega)]^2 = 4[cos( \omega)^2 + cos(4 \omega)^2 + cos(5 \omega)^2 + 2cos(\omega)cos(4\omega) + 2cos(\omega)cos(5\omega) + 2cos(4 \omega)cos(5 \omega)  ]$
Eliminate the squares via the identity $cos^2x = \frac{1 + \cos(2x)}{2}$:
$6 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[2cos(\omega)cos(4\omega) + 2cos(\omega)cos(5\omega) + 2cos(4 \omega)cos(5 \omega)]$
Next use the trig product identify on the the last 3 terms to get:
$6 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(3\omega) + cos(4 \omega) + cos(5 \omega) + cos(6 \omega) + cos(9 \omega)]$
From the initial observations $cos(3 \omega) + cos(6 \omega) + cos(9 \omega) = -\frac{1}{2}$ 
We get:
$4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(4 \omega) + cos(5 \omega)]$
Now back to the original equation add in the -2x term:
$4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(4 \omega) + cos(5 \omega)] - 2[cos(\omega) + cos(4 \omega) + cos(5 \omega)]= 
4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega) + cos(\omega) + cos(4 \omega) + cos(5 \omega)]$
But from our original observations the second sum = $\frac{1}{2}$
So we get $4 + 2[\frac{1}{2}] = 5$ which is the desired result.
