Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$ 
Prove:
  $$\begin{align}
\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt]
\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7} &=\phantom{-}24 \tag{2}
\end{align}$$


you do need to know how to work with symmetric polynomials. The basic fact is that the roots of
$$  x^3 + x^2 - 2x-1 $$ are
$$ 2 \cos \frac{2 \pi}{7}, \; \; \; 2 \cos \frac{4 \pi}{7}, \; \; \; 2 \cos \frac{6 \pi}{7}.   $$
This tells you the sum $(-1)$, the sum of pairwise products$(-2)$, and the product$(1)$. The observation goes back, essentially, to Gauss. You need to modify by factors of 2,4,8.
