Show that $\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$.

We know that $3x = 2x + x$.
Therefore,$\tan 3x = \tan (2x + x)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$.
Cross-multiplying gives:
$\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x$.
$\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x$.
Rearranging the terms,we get:
$\tan 3x - \tan 2x - \tan x = \tan 3x \tan 2x \tan x$.
Hence,$\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$.