Step 1:
Let $1+\sqrt 3i=r(\cos \theta+i\sin \theta)$
Equating the real and imaginary parts separately
$r\cos\theta=1$
$r\sin\theta=\sqrt 3$
Squaring and adding we get
$r^2=4$
$r=2$
Step 2:
Also $\alpha=\tan^{-1}\sqrt 3=\large\frac{\pi}{3}$
Since $1+\sqrt 3i$ is represented by a point in quadrant 1.
$\theta=\alpha=\large\frac{\pi}{3}$
Therefore $1+\sqrt 3i=2(\cos \large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})$
$1-\sqrt 3i=1+\sqrt 3i$
$\qquad\;\;\;\;\;=2(\cos \large\frac{\pi}{3}$$-i\sin\large\frac{\pi}{3})$
Step 3:
Now LHS=$(1+i\sqrt 3)^n+(1-i\sqrt 3)^n$
$\qquad\quad\;\;=[2(\cos\large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})]^n+$$[2(\cos\large\frac{\pi}{3}$$-i\sin\large\frac{\pi}{3})]^n$
$\qquad\quad\;\;=2^n[\cos\large\frac{n\pi}{3}$$+i\sin\large\frac{n\pi}{3}$$+\cos\large\frac{n\pi}{3}$$-i\sin\large\frac{n\pi}{3}$
$\qquad\quad\;\;=2^n.2\cos\large\frac{n\pi}{3}$
$\qquad\quad\;\;=2^{n+1}\cos\large\frac{n\pi}{3}$
$\qquad\quad\;\;$=RHS
Hence proved.