Finding The Actual Roots Might Be Painful 1

Considering the equation $AX^{3}+BX^{2}+CX+D=0$ given that $\text{A=1, B=0, C=-1, D=-1}.$ Using a vieta's theorem: $\alpha + \beta + \gamma= \frac {-B}{A}=0$ $\alpha\beta + \beta\gamma + \alpha\gamma= \frac {C}{A}=-1$ $\alpha\beta\gamma= \frac {-D}{A}=1$ . Then: $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ $\Rightarrow \frac{(1+\alpha)(1-\beta)(1-\gamma) + (1+\beta)(1-\alpha)(1-\gamma) + (1+\gamma)(1-\alpha)(1-\beta)}{(1-\alpha)(1-\beta)(1-\gamma)}$ $\Rightarrow \frac{3-(\alpha+\beta+\gamma) - (\alpha\beta + \beta\gamma + \alpha\gamma) + 3(\alpha\beta\gamma)}{1-(\alpha+\beta+\gamma) + (\alpha\beta + \beta\gamma + \alpha\gamma)-(\alpha\beta\gamma)}$ Substitued: $\Rightarrow \frac{3-0-(-1)+3(1)}{(1-0+(-1)-1)} =\boxed{-7}$

$\begin{aligned} 1+x&={x}^3\\{x}^3-1 &=x\\\left(x-1\right)\left({x}^2+x+1\right)&=x\\1-x &=\frac{-x}{{x}^2+1+x}=\frac{-x}{{x}^2+{x}^3}=\frac{-x}{{x}^2\left(1+x\right)}=-\frac{1}{{x}^4}\\\frac{1+x}{1-x}&=-{x}^7\\{x}^6&={x}^2+2x+1\\{x}^7&={x}^3+2{x}^2+x\\\sum \frac{1+x}{1-x}&=-\sum {x}^7=-\sum \left({x}^3+2{x}^2+x\right) \\&=-\left[{\alpha}^3+{\beta}^3+{\gamma}^3+2\left({\alpha}^2+{\beta}^2+{\gamma}^2\right)+\alpha+\beta+\gamma\right]\\&=-\left[3\alpha\beta\gamma+2\left(-2\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)\right)\right]\\&=-\left(3+4\right)\\&\huge\color{#3D99F6}{=\boxed{-7}}\end{aligned}$

replace x by (x-1)/(x+1) and use vieta's to find sum of roots.