So, considering LHS:
\[b\text{ }\left( c\text{ }cos\text{ }A\text{ }\text{ }a\text{ }cos\text{ }C \right)\]
\[bc\text{ }cos\text{ }A\text{ }and\text{ }ab\text{ }cos\text{ }C\]are L.H.S, which can be obtained from cosine formulae.
Using cosine formula:
\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]
\[bc\text{ }cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2\text{ }\ldots \text{ }\left( i \right)\]
Or,
\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]
\[ab\text{ }cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2\text{ }\ldots \text{ }\left( ii \right)\]
Subtracting equation (i) and (ii) we get,
\[bc\text{ }cos\text{ }A\text{ }\text{ }ab\text{ }cos\text{ }C\]
\[=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2\text{ }\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2\]
So,
\[=\text{ }{{c}^{2}}~\text{ }{{a}^{2}}\]
\[\therefore b\text{ }\left( c\text{ }cos\text{ }A\text{ }\text{ }a\text{ }cos\text{ }C \right)\text{ }=\text{ }{{c}^{2}}~\text{ }{{a}^{2}}\]
Hence proved.