Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\cos^2 40°}{\cos^2 20°.\cos^2 40°.\cos^2 80°}$$.
I got paused here. Please help to prove this..

A clean start, not remotely obvious ( but easy enough to prove ), the roots of
$$  x^3 - 3 x + 1 $$ are
$$ 2 \cos \frac{2 \pi}{9}, \; \;   2 \cos \frac{4 \pi}{9}, \; \;  2 \cos \frac{8 \pi}{9}.  $$
From page 174 in Reuschle (1875). The method used is due to Gauss. 
I learned today that, in 1933, D. H. Lehmer published a two-page proof that $2 \cos (2k\pi/n)$ is always an algebraic integer. Mentioned in an article in the January (M. A. A.) Monthly by Y. Z. Gurtas. 
We immediately get that the roots of 
$$ 8 x^3 - 6 x + 1, $$ therefore
$$  x^3 - \frac{3}{4} x + \frac{1}{8} $$ are
$$  \cos \frac{2 \pi}{9}, \; \;    \cos \frac{4 \pi}{9}, \; \;   \cos \frac{8 \pi}{9}.  $$
Compare with the three requested proofs in the answer by Michael R.
Proof comes by taking $\omega$ as a primitive ninth root of unity, then taking
$$  x = \omega + \frac{1}{\omega}. $$
Primitive means
$$ \omega \neq 1, \; \; \omega^3 \neq 1, \omega^9 = 1 $$
Then $$ x^3 = \omega^3 + 3 \omega + \frac{3}{\omega} + \frac{1}{\omega^3}, $$
$$ x^3 - 3 x + 1 = \omega^3 + 1 + \frac{1}{\omega^3},  $$
$$ \omega^3 (x^3 - 3 x + 1) = \omega^6 + \omega^3 + 1,  $$
$$ (\omega^3 - 1)\omega^3 (x^3 - 3 x + 1) = (\omega^3 - 1)(\omega^6 + \omega^3 + 1) = \omega^9 - 1 = 0.  $$
These can be combined with double angle formulas/half angle formulas.

