We know that,

\[cos\text{ }3\theta \text{ }=\text{ }4co{{s}^{3}}\theta \text{ }\text{ }3cos\theta \]

SO, \[4\text{ }co{{s}^{3}}\theta \text{ }=\text{ }cos3\theta \text{ }+\text{ }3cos\theta \]

\[co{{s}^{3}}~\theta \text{ }=\text{ }\left[ cos3\theta \text{ }+\text{ }3cos\theta  \right]/4\]…… (i)

Similarly,

\[sin\text{ }3\theta \text{ }=\text{ }3sin\text{ }\theta \text{ }\text{ }4si{{n}^{3}}~\theta \]

\[4\text{ }si{{n}^{3}}\theta \text{ }=\text{ }3sin\theta \text{ }\text{ }sin~3\theta \]

\[si{{n}^{3}}\theta \text{ }=\text{ }\left[ 3sin\theta \text{ }\text{ }sin~3\theta  \right]/4\]…….. (ii)

Now,

Let us consider LHS:

\[co{{s}^{3}}~x\text{ }sin\text{ }3x\text{ }+\text{ }si{{n}^{3}}~x\text{ }cos\text{ }3x\]

Substituting the values from equation (i) and (ii), we get

\[co{{s}^{3}}~x\text{ }sin\text{ }3x\text{ }+\text{ }si{{n}^{3}}~x\text{ }cos\text{ }3x\text{ }=\text{ }\left( cos\text{ }3x\text{ }+\text{ }3\text{ }cos\text{ }x \right)/4\text{ }sin\text{ }3x\text{ }+\text{ }\left( 3sin\text{ }x\text{ }\text{ }sin\text{ }3x \right)/4\text{ }cos\text{ }3x\]

\[=\text{ }1/4\text{ }\left( sin\text{ }3x\text{ }cos\text{ }3x\text{ }+\text{ }3\text{ }sin\text{ }3x\text{ }cox\text{ }x\text{ }+\text{ }3sin\text{ }x\text{ }cos\text{ }3x\text{ }\text{ }sin\text{ }3x\text{ }cos\text{ }3x \right)\]

\[=\text{ }1/4\text{ }\left( 3\left( sin\text{ }3x\text{ }cos\text{ }x\text{ }+\text{ }sin\text{ }x\text{ }cos\text{ }3x \right)\text{ }+\text{ }0 \right)\]

\[=\text{ }1/4\text{ }\left( 3\text{ }sin\text{ }\left( 3x\text{ }+\text{ }x \right) \right)\]

(We know, sin(x + y) = sin x cos y + cos x sin y)

\[=\text{ }3/4\text{ }sin\text{ }4x\]

= RHS

Hence proved.