Take LHS:

\[co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x\]

From formula, \[{{\left( a\text{ }+\text{ }b \right)}^{~2}}~=\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\]

\[{{a}^{3}}~\text{ }{{b}^{3}}~=\text{ }\left( a\text{ }\text{ }b \right)\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }ab)\]

So,

\[co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x\text{ }=\text{ }{{(co{{s}^{2}}~x)}^{3}}~\text{ }{{(si{{n}^{2}}~x)}^{3}}\]

\[=\text{ }(co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x)\text{ }(co{{s}^{4}}~x\text{ }+\text{ }si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x)\]

Form formula, \[\text{ }cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\]

So,

\[=\text{ }cos\text{ }2x\text{ }[{{(co{{s}^{2}}~x)}^{~2}}~+\text{ }{{(si{{n}^{2}}~x)}^{~2}}~+\text{ }2\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x\text{ }\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x]\]

\[=\text{ }cos\text{ }2x\text{ }[{{(co{{s}^{2}}~x)}^{~2}}~+\text{ }{{(si{{n}^{2}}~x)}^{~2}}~\text{ }1/4\text{ }\times \text{ }4\text{ }co{{s}^{2}}~x\text{ }si{{n}^{2}}~x]\]

From formula, \[si{{n}^{2}}~x\text{ }+\text{ }co{{s}^{2}}~x\text{ }=\text{ }1\]

So,

\[=\text{ }cos\text{ }2x\text{ }[{{\left( 1 \right)}^{2}}~\text{ }1/4\text{ }\times \text{ }{{\left( 2\text{ }cos\text{ }x\text{ }sin\text{ }x \right)}^{~2}}]\]

So,

\[=\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }\times \text{ }{{\left( sin\text{ }2x \right)}^{~2}}]\]

\[=\text{ }cos\text{ }2x\text{ }[1\text{ }\text{ }1/4\text{ }\times \text{ }si{{n}^{2}}~2x]\]

= RHS

Hence proved.