tan 20° tan 40° tan 80° and tan²20° + tan²40° + tan²80°

1. With $t=\tan\theta$,
$$\tan(60^\circ+\theta)\tan(60^\circ-\theta)=\frac{(\sqrt3+t)(\sqrt3-t)}{(1-\sqrt3t)(1+\sqrt3t)}=\frac{3-t^2}{1-3t^2},$$
so the product is $\dfrac{t(3-t^2)}{1-3t^2}=\tan3\theta$ (the triple-angle formula).

2. For $\theta=20^\circ$: $60^\circ-20^\circ=40^\circ$ and $60^\circ+20^\circ=80^\circ$, so the product is $\tan60^\circ=\sqrt3$.

3. The equation $\tan3\theta=\sqrt3$ means $\frac{3t-t^3}{1-3t^2}=\sqrt3$, i.e.
$$t^3-3\sqrt3\,t^2-3t+\sqrt3=0.$$
Its solutions $t=\tan\theta$ include $\theta=20^\circ,\ 80^\circ,\ 140^\circ$ (since $3\theta=60^\circ,240^\circ,420^\circ$, all with tangent $\sqrt3$). These give the three roots $\tan20^\circ$, $\tan80^\circ$, $\tan140^\circ=-\tan40^\circ$. By Vieta, with $e_1=3\sqrt3$ and $e_2=-3$:
$$\tan^220^\circ+\tan^280^\circ+\tan^2(-40^\circ)=e_1^2-2e_2=27+6=33.$$
Since $\tan^2(-40^\circ)=\tan^240^\circ$, this is the required sum. $\blacksquare$

Lesson. When several trigonometric values share an angle equation ($\tan3\theta=\text{const}$), they are the roots of one polynomial: use Vieta for their symmetric functions.