Prove that,
$(\sqrt{3}+ 1) (3-\cot 30^{\circ})=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$

(A) We know the trigonometric values: $\tan 60^{\circ} = \sqrt{3}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 30^{\circ} = \sqrt{3}$.
First,evaluate the $R$.$H$.$S$.:
$\text{R.H.S.} = \tan^{3} 60^{\circ} - 2 \sin 60^{\circ} = (\sqrt{3})^{3} - 2 \left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}$.
Next,evaluate the $L$.$H$.$S$.:
$\text{L.H.S.} = (\sqrt{3} + 1)(3 - \cot 30^{\circ}) = (\sqrt{3} + 1)(3 - \sqrt{3})$.
Factor out $\sqrt{3}$ from the second bracket:
$\text{L.H.S.} = (\sqrt{3} + 1) \sqrt{3}(\sqrt{3} - 1) = \sqrt{3}(\sqrt{3} + 1)(\sqrt{3} - 1)$.
Using the identity $(a+b)(a-b) = a^2 - b^2$:
$\text{L.H.S.} = \sqrt{3}((\sqrt{3})^2 - 1^2) = \sqrt{3}(3 - 1) = \sqrt{3}(2) = 2\sqrt{3}$.
Since $\text{L.H.S.} = \text{R.H.S.}$,the identity is proved.