Motivation: find a polynomial with roots $\cos(2\pi/7)$, $\cos(4\pi/7)$, $\cos(6\pi/7)$. I provide some steps. If you are stuck, feel free to ask although I have included as many details I could.
Consider the seventh roots of unity $f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ with roots $x = \cos(2k\pi/7) + i\sin(2k\pi/7)$ where $k\in \{1,2,3,4,5,6\}$.
It follows that setting $t= x+\frac1{x}$ gives us that $f(x) = x^3(x^3 + \frac1{x^3} + x^2 + \frac1{x^2} + x + \frac1{x} + 1)$. Use that $x^3+1/x^3 = t^3-3t$ and $x^2+1/x^2 = t^2-2$. Thus $f(x) = x^3(t^3 - 3t + t^2 - 2 + t + 1)$. So setting $f(x) = 0$ gives that $t^3+t^2-2t-1 = 0$. This polynomial in $x$ has the roots defined as above but in $t$ the roots are just $x+\frac1{x}$ which is $2\cos(2k\pi/7)$ where $k\in \{1,2,3\}$. Set $t\mapsto 2t$ to get the polynomial with the correct roots, viz. $8t^3+4t^2-4t-1=0$
Let us denote each of the cosines by $x^3,y^3,z^3$ respectively. Recall the identity $a^3+b^3+c^3 - 3abc = (a+b+c)((a+b+c)^2 - 3ab - 3bc - 3ca)$. Setting first $(a,b,c) \mapsto (x,y,z)$ and $(a,b,c) \mapsto (xy,yz,zx)$, where $x+y+z = P$ and $xy + yz + zx = Q$; we get the following two equations:
$$\begin{align}
(x^3 + y^3 + z^3) - 3(xyz) &= P(P^2 - 3Q) \\
(x^3y^3 + y^3z^3 + x^3z^3) - 3(x^2y^2z^2) &= Q(Q^2 - 3Pxyz)
\end{align}$$
All the elementric symmetric polynomials in $x^3, y^3, z^3$ can be evaluated using the polynomial mentioned in point $2$, leaving us two equations in $P$ and $Q$.
Now solve for $P$. One must be careful of what branch of the cube root is chosen for finding $xyz$ and $x^2y^2z^2$ from $x^3y^3z^3$ which is known.
- Rendering the result explicitly, we have
$$x^3+y^3+z^3=-(1/2)$$
$$xyz=1/2^*$$
$$x^3y^3+x^3z^3+y^3z^3=-(1/2)$$
$$x^2y^2z^2=1/4^*$$
(*-- this comes from the real cube-root branch of $x^3y^3z^3=1/8$, which is chosen to assure consistency with the real values assumed in the original sum.)
So
$$Q=\frac{P^3+2}{3P}$$
$$-(5/4)=\dfrac{P^3+2}{3P}\left(\left(\dfrac{P^3+2}{3P}\right)^2-\dfrac{3P}2\right)\implies 4P^9-30P^6+75P^3+32=0$$
This may be solved by the Cardano method for $P^3$, giving only one real value for $P^3$, viz.
$$P^3=\dfrac{5-3\sqrt[3]7}{2}$$
from which the claimed result follows.
