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 :)
 A: 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*}
A: Nice solutions guys, I have got a different solution:
If $\{x_i\}$ are roots of $P(x)=0$, then $\left\{\frac{1+x_i}{2-x_i}\right\}$ are roots of $P\left(\frac{2x-1}{x+1}\right)=0$
So $$(2x-1)^3+(2x-1)^2(x+1)+(x+1)^3=0$$
So $$13x^3-9x^2+6x+1=0$$
Hence the result ; $\boxed{\frac 9{13}}$
A: Notice, we have $x^3+x^2+1=0$
$$\alpha+\beta+\gamma=-1$$
$$\alpha\beta+\beta\gamma+\alpha\gamma=0$$
$$\alpha\beta\gamma=-1$$
 $$\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma }$$   $$=\left(\frac{1+\alpha }{2-\alpha }+1\right)+\left(\frac{1+\beta }{2-\beta }+1\right)+\left(\frac{1+\gamma }{2-\gamma }+1\right)-3$$
$$=\left(\frac{3 }{2-\alpha }\right)+\left(\frac{3 }{2-\beta }\right)+\left(\frac{3 }{2-\gamma }\right)-3$$
$$=\frac{3[(2-\beta)(2-\gamma)+(2-\alpha)(2-\gamma)+(2-\beta)(2-\gamma)]}{(2-\alpha)(2-\beta)(2-\gamma)}-3$$
$$=\frac{3[12-4(\alpha+\beta+\gamma)+(\alpha\beta+\beta\gamma+\alpha\gamma)]}{8-4(\alpha+\beta+\gamma)+2(\alpha\beta+\beta\gamma+\alpha\gamma)-\alpha\beta\gamma}-3$$
$$=\frac{3[12-4(-1)+0]}{8-4(-1)+2(0)-(-1)}-3=\frac{48}{13}-3$$$$=\frac{9}{13}$$
