Take RHS:
\[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\]
By expanding we get,
\[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\text{ }=\text{ }4\text{ }[{{(co{{s}^{2}}~x)}^{3}}~\text{ }{{(si{{n}^{2}}~x)}^{3}}]\]
\[=\text{ }4\text{ }(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]
From Formula, \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\]
So,
\[=\text{ }4\text{ }cos\text{ }2x\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x\text{ }\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{(co{{s}^{2}}~x)}^{2}}~+\text{ }{{(si{{n}^{2}}~x)}^{2}}~+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x\text{ }\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x]\]
We know that, \[\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\text{ }=\text{ }{{\left( a\text{ }+\text{ }b \right)}^{2}}\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{\left( 1 \right)}^{2}}~\text{ }1/4\text{ }(4\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)]\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[{{\left( 1 \right)}^{2}}~\text{ }1/4\text{ }{{\left( 2\text{ }cos\text{ }x\text{ }sin\text{ }x \right)}^{2}}]\]
From formula, \[sin\text{ }2x\text{ }=\text{ }2sin\text{ }x\text{ }cos\text{ }x\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[({{1}^{2}})\text{ }\text{ }1/4\text{ }{{\left( sin\text{ }2x \right)}^{2}}]\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }(1\text{ }\text{ }1/4\text{ }si{{n}^{2}}~2x)\]
From formula, \[\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\text{ }\text{ }co{{s}^{2}}~x\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }(1\text{ }\text{ }co{{s}^{2}}~2x)]\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }+\text{ }1/4\text{ }co{{s}^{2}}~2x]\]
\[=\text{ }4\text{ }cos\text{ }2x\text{ }[3/4\text{ }+\text{ }1/4\text{ }co{{s}^{2}}~2x]\]
\[=\text{ }4\text{ }(3/4\text{ }cos\text{ }2x\text{ }+\text{ }1/4\text{ }co{{s}^{3}}~2x)\]
\[=\text{ }3\text{ }cos\text{ }2x\text{ }+\text{ }co{{s}^{3}}~2x\]
\[=\text{ }co{{s}^{3}}~2x\text{ }+\text{ }3\text{ }cos\text{ }2x\]
= LHS
Hence proved.