#### Answer

$\displaystyle \frac{3\tan x-\tan^{3}x}{1-\tan^{3}x}$

#### Work Step by Step

$\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$ \displaystyle \tan 2A=\frac{2\tan A}{1-\tan^{2}A}$
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$\tan 3x=\tan(x+2x)$
$=\displaystyle \frac{\tan x+\tan 2x}{1-\tan x\tan 2x}$
$=\displaystyle \frac{\tan x+\frac{2\tan x}{1-\tan^{2}x}}{1-\tan x\cdot\frac{2\tan x}{1-\tan^{2}x}}=\frac{\frac{\tan x(1-\tan^{2}x)+2\tan x}{1-\tan^{2}x}}{\frac{(1-\tan^{2}x)-2\tan^{2}x}{1-\tan^{2}x}}$
$=\displaystyle \frac{\tan x(1-\tan^{2}x)+2\tan x}{(1-\tan^{2}x)-2\tan^{2}x}$
$=\displaystyle \frac{3\tan x-\tan^{3}x}{1-\tan^{3}x}$