$x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ 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!

HINT:
The polynomial $4x^{3}+ 2x^{2}- 7x- 5$ factors as $(x + 1) (4 x^2 - 2 x - 5)$. 
So we need to prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a root of $(4 x^2 - 2 x - 5)$. What is the other root? It is $x_1=\cos \frac{4\pi }{21}+ \cos \frac{16\pi }{21}+ \cos\frac{20\pi }{21}$. 
How does this work? Note that we are dealing with conjugate elements $\cos\frac{2 k \pi}{21}$, where $k \in (\mathbb{Z}/21)^{\times}/{\pm 1}$. The Galois group of the extension generated by them is isomorphic to $G\colon =(\mathbb{Z}/21)^{\times}/{\pm 1}$, a cyclic group of order $6$ (note that $(\mathbb{Z}/21)^{\times}$ itself is not cyclic). The transformation $\rho_{a}$ maps $\cos \frac{2 k \pi}{21}$ to $\cos \frac{2 a k \pi}{21}$. 
Now $G$ has a unique subgroup of order $3$ with representatives ${1,4,5}$ (the squares of elements of $G$). ${2,8,10}$ is the coset of $2$. Now it's not that hard to see that $x_0$, $x_1$ are conjugate algebraic numbers. 
It's not that hard to check that $x_0+x_1=-\frac{1}{2}$, using the cyclotomic polynomial $\Phi_{21}(x)$ to produce an equation of degree $6$ with roots $\cos\frac{2 k \pi}{21}$, $k \in {1,2,4,5,8,10}$. It seems like an interesting exercise to check that $x_0 \cdot x_1=-\frac{5}{4}$.
ADDED: One can check that the sum $\sum_{a \in \mathbb{Z}/21)^{\times}} e^{\frac{2 a\pi }{21}}$ is just $2(x_0+x_1)$. But the sum of the roots of the cyclotomic polynomial $\Phi_{21}(z)$ is $-1$. Now for $x_0 \cdot x_1$ one can do calculations by hand using formulas for $\cos \alpha \cdot \cos \beta$.
Comment: The method in the other answer of @Will Jagy: works fine when we need to check that the expression is a root of a given polynomial, it's all automatic. 
However, when one has a number like $\sum_{a \in H} e^{{2 a \pi}{n}}$ for a subgroup $H$ of $(\mathbb{Z}/n)^{\times}$,  one finds right away the conjugates ( in fact one can do this for any elements of a cyclotomic field), then, using sufficient precision, select the distinct ones, and finds the minimal polynomial(now preferably to work with algebraic  integers).
