trigonometry - Prove that $cos^6 beta - sin^6 beta=frac {1}{16} (15 cos 2beta + cos 6beta)$.

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.

Answer

$$(\cos^2\beta)^3 -(\sin^2\beta)^3=(\cos^2\beta-\sin^2\beta)(\cos^4\beta +\sin^4\beta +\cos^2\beta\sin^2\beta)$$
$$=\cos2\beta(1-\sin^2\beta\cos^2\beta)$$
$$=\cos2\beta\left(1-\frac{\sin^2 2\beta}{4}\right)$$
$$=\cos2\beta\left(1+\frac{\cos 4\beta -1}{8}\right)$$
Now use $2\cos A.\cos B=\cos(A+B) +\cos(A-B)$
$$=\frac18\left(7\cos2\beta +\cos 2\beta\cos 4\beta\right)$$
$$=\frac18\left(7\cos 2\beta +\frac{\cos 6\beta + \cos 2\beta}{2}\right)$$
$$=\frac{1}{16}\left(15\cos 2\beta + \cos 6 \beta\right)$$