We have

$(x+1)^6 + (x-1)^6 = 0$

$\therefore (x+1)^6 = - (x-1)^6$

$\therefore \left ( \dfrac {x+1}{x-1} \right )^6 = -1 = \cos \pi + i \sin \pi = \cos (2k+1) \pi + i \sin (2k + 1) \pi$

$\therefore \dfrac {x+1}{x-1} = \cos (2k+1) \dfrac {\pi}{6} + i \sin (2k+1) \dfrac {\pi}{6}$

Let $(2k+1) \dfrac {\pi}{6} = \theta $

$\therefore \dfrac {x+1}{x-1} = \cos \theta + i \sin \theta$

Applying componendo and dividendo,

$\therefore \dfrac{x}{-1} = \dfrac {1+ \cos \theta + i \sin \theta}{1- \cos \theta - i \sin \theta} = \dfrac {2 \cos ^2 \left ( \frac {\theta}{2} \right ) + 2i \sin \left ( \frac {\theta}{2} \right )\cos \left ( \frac {\theta}{2} \right )} {2 \sin^2 \left ( \frac {\theta}{2} \right ) - 2i \sin \left ( \frac {\theta}{2} \right ) \cos \left ( \frac {\theta}{2} \right )}$

$\frac {x}{1} = \cot \left ( \dfrac {\theta}{2} \right ) \dfrac {\cos \left ( \frac {\theta}{2} \right ) + i \sin \left ( \frac {\theta}{2} \right )}{- \sin \left ( \frac {\theta}{2} \right ) + i \cos \left ( \frac {\theta}{2} \right )} $

$ \therefore \dfrac {x}{1} = \dfrac {e^{\frac {i\theta}{2}}}{e^{i \left ( \frac {\theta}{2} + \frac {\pi}{2} \right )}} \cot \left ( \dfrac {\theta}{2} \right ) = \cot \left ( \dfrac {\theta}{2} \right )e^{-i\frac {\pi}{2}}$

$ = - i \cot \left ( \dfrac{\theta}{2} \right )$

$\therefore (x+1)^6 + (x-1)^6 = 0 = - i \cot \left [ \dfrac {(2k+1)\pi}{12} \right ]$

Hence Proved....