According to sine rule,

now,

\[~c\text{ }=\text{ }k\text{ }sin\text{ }C\]

same way,

\[a\text{ }=\text{ }k\text{ }sin\text{ }A\]

also,

\[b\text{ }=\text{ }k\text{ }sin\text{ }B\]

As we know,

Now let us consider RHS:

\[\left( {{b}^{2}}~\text{ }{{c}^{2}} \right)\text{ }sin\text{ }A\text{ }\ldots \]

Substituting the values of b and c we have,

\[\left( {{b}^{2}}~\text{ }{{c}^{2}} \right)\text{ }sin\text{ }A\text{ }=\text{ }\left[ {{\left( k\text{ }sin\text{ }B \right)}^{2}}~\text{ }{{\left( \text{ }k\text{ }sin\text{ }C \right)}^{2}} \right]\text{ }sin\text{ }A\]

\[=\text{ }{{k}^{2~}}\left( si{{n}^{2}}~B\text{ }\text{ }si{{n}^{2}}~C \right)\text{ }sin\text{ }A\ldots \ldots \ldots .\text{ }\left( i \right)\]

We know,

\[Si{{n}^{2}}~B\text{ }\text{ }si{{n}^{2}}~C\text{ }=\text{ }sin\text{ }\left( B\text{ }+\text{ }C \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right),\]

Substituting the above values in equation (i), we have

\[~=\text{ }{{k}^{2~}}\left( sin\text{ }\left( B\text{ }+\text{ }C \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\text{ }sin\text{ }A\]

\[[since,\text{ }\pi \text{ }=\text{ }A\text{ }+\text{ }B\text{ }+\text{ }C\Rightarrow ~B\text{ }+\text{ }C\text{ }=\text{ }\pi \text{ }A]\]

\[~=\text{ }{{k}^{2~}}\left( sin\text{ }\left( \pi \text{ }A \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\text{ }sin\text{ }A\]

\[=\text{ }{{k}^{2~}}\left( sin\text{ }\left( A \right)\text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\text{ }sin\text{ }A\]

\[\left[ since,\text{ }sin\text{ }\left( \pi \text{ }\text{ }\theta  \right)\text{ }=\text{ }sin\text{ }\theta  \right]\]

\[=\text{ }\left( k\text{ }sin\text{ }\left( A \right) \right)\left( \text{ }sin\text{ }\left( B\text{ }\text{ }C \right) \right)\text{ }\left( k\text{ }sin\text{ }A \right)\]

From sine rule, \[a\text{ }=\text{ }k\text{ }sin\text{ }A,\]

so the above equation becomes,

\[=\text{ }{{a}^{2}}~sin\text{ }\left( B\text{ }\text{ }C \right)\]

\[=\text{ }RHS\]

Hence proved.