If $\alpha, \beta, \gamma$ solutions for the equation If $\alpha, \beta, \gamma$ solutions for the equation
$$x^{3}+x^{2}+1=0$$
Then
$$\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma }=??$$
I know that the answer is $\frac{9}{13}$, It's just for sharing a new ideas, thanks :)

Using the root-coefficient relationship, we have the following:
$$\alpha + \beta + \gamma = -1,$$
$$\alpha\beta\gamma = -1,$$
$$\alpha\beta + \beta\gamma + \gamma\alpha = 0.$$
The sum in question can be evaluated easily now.
\begin{align*}
\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma } &= \frac{3(\alpha\beta + \beta\gamma + \gamma\alpha - \alpha\beta\gamma) -12}{\alpha\beta\gamma -2 (\alpha\beta +\beta\gamma + \gamma\alpha) + 4(\alpha + \beta + \gamma) -8} \\
&= \frac{3(0 + 1) -12}{-1 -2 (0) + 4(-1) -8} \\
&= \frac{9}{13}
\end{align*}
