<s>[INST] Prove that $\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$. 
Prove the given trigonometric identity:
  $$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$

My Approach:
\begin{align*}
\text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\
&=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\\
&=\cos^32\beta+3\cos^2\beta\cdot\sin^2\beta\cdot\cos2\beta
\end{align*}
Please help me to continue further.
 [/INST] $$\cos^{6} (x)-\sin^{6}(x)=(\cos^{2}(x)-\sin^{2}(x))(\cos^{4} (x)+\cos^{2}(x)\sin^{2}(x)+\sin^{4}(x))$$
$$a^{2}+b^{2}=(a+b)-2ab.$$
$$=\cos (2 x)(1-\cos^{2} (x)\sin^{2}(x))$$
$$=\cos 2x-\dfrac{1}{2}\cos (2x)\sin^{2}(2x)$$
$$=\dfrac{\cos 2x(1+\cos^{2}(2x))}{2}....؟$$
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