Prove the following:-
\[\left( \sqrt{3}+1 \right)\left( 3-\cot {{30}^{\circ }} \right)={{\tan }^{3}}{{60}^{\circ }}-2\sin {{60}^{\circ }}\]

Hint: This solution can be done by either evaluating the left hand side that is the L.H.S. or by evaluating the right hand side that is the R.H.S.
Now, important results that can be used in solving this question are
 \[\tan {{60}^{\circ }}=\sqrt{3}\]
\[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
\[\cot {{30}^{\circ }}=\sqrt{3}\]

Complete step-by-step answer:
As mentioned in the question, we have to solve the question and prove the equation by either evaluating the left hand side that is the L.H.S. or by evaluating the right hand side that is the R.H.S.
Now, if we evaluate the left hand side or the L.H.S., we can solve as follows
\[\begin{align}
  & =\left( \sqrt{3}+1 \right)\left( 3-\cot {{30}^{\circ }} \right) \\
 & =\left( \sqrt{3}+1 \right)\left( 3-\sqrt{3} \right) \\
 & =3\sqrt{3}+3-3-\sqrt{3} \\
 & =3\sqrt{3}-\sqrt{3} \\
\end{align}\]
Now, we know that \[\tan {{60}^{\circ }}=\sqrt{3}\] and \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\]as mentioned in the hint, then we can write the equation is as follows
\[\begin{align}
  & =3\sqrt{3}-\sqrt{3} \\
 & ={{\tan }^{3}}{{60}^{\circ }}-2\sin {{60}^{\circ }} \\
\end{align}\]
As this is equal to the right hand side or the R.H.S., hence, this is proved.

Note: The students can make an error if they don’t know the facts that are mentioned in the hint that is as follows
This solution can be done by either evaluating the left hand side that is the L.H.S. or by evaluating the right hand side that is the R.H.S.
Now, important results that can be used in solving this question are
 \[\tan {{60}^{\circ }}=\sqrt{3}\]
\[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
\[\cot {{30}^{\circ }}=\sqrt{3}\]