($x+1 )^6 + (x-1)^6\;=\; 0 \; \; \therefore (x+1 )^6 \;=\; -(x-1)^6 \\ \; \\ \; \\ \; \\ \therefore \Bigg[ \dfrac{x+1}{x-1} \Bigg]^6 \;=\; -1 \;=\; cos(2k+1)\pi + isin(2k+1)\pi \; \; \; , k=0,1,2,3,4,5 \\ \; \\ \; \\ \; \\ \therefore \dfrac{x+1}{x-1} \;=\; \Bigg[ cos(2k+1)\pi + isin(2k+1)\pi \Bigg] ^{\frac{1}{6}} \\ \; \\ \; \\ \; \\ \therefore \dfrac{x+1}{x-1} \;=\; \Bigg[ cos\Big( \dfrac{2k+1}{6}\Big)\pi + isin\Big( \dfrac{2k+1}{6}\Big)\pi \Bigg] \; = \; e^{i\Big( \dfrac{2k+1}{6}\Big)\pi} \;=\; say \; e^{i\theta} \\ \; \\ where \; \theta \;=\; \Big( \dfrac{2k+1}{6}\Big)\pi \\ \; \\ \; \\ \; \\ $

Applying componendo- dividend, $ \therefore \dfrac{x+1+x-1}{x+1-x+1} \;=\; \dfrac{e^{i\theta} +1}{e^{i\theta}-1} \\ \; \\ \; \\ \therefore z\;=\; \dfrac{e^{i\theta}}{1-e^{i\theta}} \;=\; \dfrac{cos\theta+isin\theta}{1-cos\theta-isin\theta} \\ \; \\ \; \\ \therefore z \;=\; \dfrac{(cos\theta+isin\theta)[(1-cos\theta)+isin\theta]}{[1-cos\theta-isin\theta][(1-cos\theta)+isin\theta]} \\ \; \\ \; \\ \therefore z\;=\; \dfrac{cos\theta - cos^2 \theta + isin\theta -isin\theta cos\theta + isin\theta cos \theta + i^2 sin^2 \theta }{(1-cos\theta)^2 - i^2 sin^\theta} \\ \; \\ \; \\ = \dfrac{cos\theta -1+ isin\theta}{ 1+1-2cos\theta} \;=\; \dfrac{-(1-cos\theta)+isin\theta}{2(1-cos\theta)} \\ \; \\ \; \\ \; \\ \therefore z \;= \; \dfrac{-1}{2} \;+\; \dfrac{isin\theta}{ 2(1-cos\theta)} \;=\; \dfrac{-1}{2} \;+\; \dfrac{2i \; sin\theta/2 \; cos\theta/2}{ 2 \cdot 2 sin^2 \theta/2} \\ \; \\ \; \\ z \;=\; \dfrac{-1}{2} \;+\; \dfrac{i}{2} cot \dfrac{\theta}{2} \; \; \; \; \ldots where \; \theta \;=\; \dfrac{2n\pi}{3} $

Hence Proved.