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!
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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).
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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.
There is a modern treatment of this material (and how Gauss did it) in Galois Theory by David A. Cox. Actually, the publisher will sell individual chapters, and the cyclotomic extension stuff is all in one chapter(9). Having done this question, I finally understand what Reuschle (1875) has for "denominator" $pq$ from separate primes $p,q \; \; .$
Let $$ x = e^{2 \pi i / 21} $$ be a primitive 21'st root of unity, so that $$ x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1 =0$$
Because, you see, $$ x^{21} - 1 = \left( x-1 \right) \left( x^2 + x + 1 \right) \left( x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \right) \left( x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1\right) $$
Next, take $$ y = x + \frac{1}{x} + x^4 + \frac{1}{x^4} + x^{16} + \frac{1}{x^{16}} = x + \frac{1}{x} + x^4 + \frac{1}{x^4} + \frac{1}{x^{5}} + x^5 $$
Calculate $$ y^3 + y^2 - 7 y - 10 $$ with the understanding about $x$ above. I was surprised at not finding this in Reuschle, but it is reducible, $(y+2)(y^2 - y - 5),$ and it is the quadratic factor on page 204 of Reuschle. Good to know.
Next Day: I did the long way already. This time I am doing the easier way, $$ x^{10}(y^2 - y - 5) = x^{20} + 2x^{19} + x^{18} + 2x^{16} + x^{15} + x^{14} + 2x^{13} + x^{12} + x^{11} + x^{10} + x^9 + x^8 + 2x^7 + x^6 + x^5 + 2x^4 + x^2 + 2x + 1 $$
However, this $ x^{10}(y^2 - y - 5)$ is just $$ \left( x^8 + 3 x^7 + 4 x^6 + 3 x^5 + 3 x^4 + 3 x^3 + 4 x^2 + 3 x + 1 \right) \left(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1 \right) $$ and is actually zero.
Here is the way the conclusion is given in Reuschle (1875) page 204. The main variables are $\omega$ and $\varpi \; \; ,$ which Latex calls varpi.
ORIGINAL: Alright, multiply $ y^3 + y^2 - 7 y - 10 $ by $x^{15}$ and reduce by the degree 12 polynomial, it comes out to $$ 9 + 9 x + 10 x^2 + 9 x^3 + 10 x^4 + 10 x^5 + 9 x^6 + 10 x^7 + 9 x^8 + 9 x^9 + 10 x^{10} + 10 x^{11} + 9 x^{12} + 10 x^{13} + 10 x^{14} + 8 x^{15} + 10 x^{16} + + 10 x^{17} + 9 x^{18} + 10 x^{19} + 10 x^{20} \; \; . $$ This is, however, $$ \left( 9 + 18 x + 28 x^2 + 28 x^3 + 29 x^4 + 29 x^5 + 29 x^6 + 20 x^7 + 10 x^8 \right) \left(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1 \right) $$ and is actually zero.