<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] Using Euler Identity $$e^{ix}=\cos x+i\sin x,$$
$2\cos x=e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$
$$64(\cos^6\beta-\sin^6\beta)=(2\cos\beta)^6+(2i\sin\beta)^6$$
$$=(e^{i\beta}+e^{-i\beta})^6+(e^{i\beta}-e^{-i\beta})^6$$
$$=2\left\{e^{i6\beta}+e^{-i(6\beta)}+\binom62\left(e^{i6\beta}\right)^{(6-2-2)}+\binom64(e^{i\beta})^{-2}\right\}$$
$$=2\cos6\beta+2\cdot15\cos2\beta$$
</s>