If \(\alpha, \beta\) and \(\gamma\) are roots of \(x^{3}-2 x+1=0\), then the value of \(\sum\left(\frac{1}{\alpha+\beta-\gamma}\right)\) is

Given cubic
\(x^{3}-2 x+1=0,(\alpha, \beta, \gamma)\) are roots of this
equation.
Then, sum of roots \(\quad \Sigma \alpha=0\)
\(\Rightarrow \quad \alpha+\beta+\gamma=0\)
\(\Sigma \Sigma \alpha \beta=-2, \quad \alpha \beta \gamma=-1\)
Now, we have
\[
\begin{aligned}
\Sigma \frac{1}{\alpha+\beta-\gamma} &=\Sigma \frac{1}{-\gamma-\gamma}=-\frac{1}{2} \Sigma \frac{1}{\gamma} \\
&=-\frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right) \\
&=-\frac{1}{2}\left(\frac{\alpha \beta+\beta \gamma+\alpha \gamma}{\alpha \beta \gamma}\right)
\end{aligned}
\]
\(=-\frac{1}{2} \cdot \frac{\Sigma \alpha \beta}{\alpha \beta \gamma}=-\frac{1}{2} \cdot \frac{(-2)}{(-1)}=-1\)