If $\alpha \ne 1$ is an $n^{\text{th }}$ root of unity and $n\in N$ such that frist three terms in the expansion of $(\alpha +x{)}^n$are 1, $\overline{\alpha }$ and $\frac{(n-1)}{2n}{\overline{\alpha }}^2$ then the value of $x,$ is 

Let $\alpha =e\frac{i2r\pi }{n}$ where $0<r<n$ Then,

${\alpha }^{n-1}=e^{i2\pi r(n-1)}{n}^{\frac{-i2r\pi }{n}}=e^{\frac{-i2}{\alpha }}$and ${\alpha }^{n-2}={\overline{\alpha }}^2$

Now, 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\alpha +x{)}^n{=}^n{C}_0{\alpha }^n{x}^0{+}^n{C}_1{\alpha }^{n-1}x{+}^n{C}_2{\alpha }^{n-2}{x}^2+\dots ..\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\alpha +x{)}^n=1+n\overline{\alpha }x+\frac{n(n-1)}{2}(\overline{\alpha }{)}^2{x}^2\left[\because {\alpha }^n=1\right]\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}nx=1\text{ and }\frac{n(n-1)}{2}{x}^2=\frac{n-1}{2n}\Rightarrow x=\frac{1}{n}\end{array}$