Prove the following

RHS $=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}=(\sqrt{3})^{3}-2 \frac{\sqrt{3}}{3}=3 \sqrt{3}-\sqrt{3}=2 \sqrt{3}$

$\mathrm{LHS}=(\sqrt{3}+1)\left(3-\cot 30^{\circ}\right)=(\sqrt{3}+1)(3-\sqrt{3})$

$\left[\because \tan 60^{\circ}=\sqrt{3} \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right.$ and $\left.=(\sqrt{3}+1) \sqrt{3}(\sqrt{3}-1) \cot 30^{\circ}=\sqrt{3}\right]$

$\left.=\sqrt{3}(\sqrt{3})^{2}-1\right)=\sqrt{3}(3-1)=2 \sqrt{3}$

$\therefore \quad L H S=R H S \quad$ Hence proved.