Prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ using the roots of a polynomial I have to 

prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ by using the roots of the polynomial $x^3-21x^2+35x-7=0$ 

I tried to factor the polynomial but it didn't work and later found it cannot factorize with rationals. and I saw some similar problems in StackExchange. But the answers are very complex to me. I cannot use Euler's complex number formula since it's not in the syllabus. I do not want the exact answer but guidance to the answer.

let $t=\tan(\theta)$, we have 
\begin{eqnarray*}
\tan(7 \theta) =\frac{ 7t-35t^3+21t^5-t^7}{1-21t^2+35t^4-7t^6}.
\end{eqnarray*}
Set $\tan(7 \theta) =0$ then the polynomial 
\begin{eqnarray*}
 7t-35t^3+21t^5-t^7=0
\end{eqnarray*}
has roots $t=0, \tan( \pi/7), \cdots ,\tan( 6 \pi/7)$. So
\begin{eqnarray*}
x^3-21x^2+35x-7=0
\end{eqnarray*}
has roots $x= \tan^2(\pi/7),\tan^2(2\pi/7),\tan^2(3\pi/7)$. Now let $y=x+1$.
