2015/076) Prove that: $\tan\,x\tan(60^\circ-x)\tan(60^\circ+x) = \tan3x$

LHS = $\tan\,x\tan(60^\circ-x)\tan(60^\circ+x)$
= $\tan\,x\dfrac{\tan\,60^\circ-\tan\,x}{1+\tan\,60^\circ \tan\,x}\dfrac{\tan\,60^\circ+\tan\,x}{1-\tan\,60^\circ \tan\,x}$
=  $\tan\,x\dfrac{\tan^260^\circ-\tan^2x}{1- \tan^260^\circ \tan^2x}$
=  $\tan\,x\dfrac{3-\tan^2x}{1- 3 \tan^2x}$
=  $\dfrac{3\tan\,x-\tan^3x}{1- 3 \tan^2x}$
= $\tan 3x$ = RHS