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.

Use $$\cos2A=2\cos^2A-1=1-2\sin^2A$$  to find
$$8(\cos^6 \beta - \sin^6 \beta)=(2\cos^2\beta)^3-(2\sin^2\beta)^3=(1+\cos2\beta)^3-(1-\cos2\beta)^3=?$$
Finally use $$\cos3B=4\cos^3B-3\cos B\iff4\cos^3B=?$$
