How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$

How can I find the following product using elementary trigonometry?

$$ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$$

I have tried using a substitution, but nothing has worked.

A simpler version of my answer to Evaluate $\tan ^{2}20^{\circ}+\tan ^{2}40^{\circ}+\tan ^{2}80^{\circ}$ works here. Basically: Write $\tan 3x$ as a polynomial in $\tan x$. The equation $\tan 3x = \sqrt{3}$ has roots $x = 20, 140$, and $260$ degrees. Thus the constant term of the polynomial in $\tan x$ you get with this equation gives $$-\tan(20^{\circ})\tan(140^{\circ})\tan(260^{\circ})$$ Since $\tan$ has period $180^{\circ}$ this is the same as $$-\tan(20^{\circ})\tan(-40^{\circ})\tan(80^{\circ})$$ $$ = \tan(20^{\circ})\tan(40^{\circ})\tan(80^{\circ})$$

From Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,

$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$

Set $x=20^\circ$

See also : A trigonometric equation