# If $α, β, γ$ are roots of $x^3 – x -1 = 0$, then find the value of $\frac{1+α}{1-α} + \frac{1+β}{1-β} + \frac{1+γ}{1-γ}$.

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If $α, β, γ$ are roots of $x^3 – x -1 = 0$, then find the value of $$\frac{1+α}{1-α} + \frac{1+β}{1-β} + \frac{1+γ}{1-γ}$$
I found this question asked in a previous year competitive examination, which was multiple choice in nature, the available options to the question were:

1. $1$
2. $0$
3. $-7$
4. $-5$

Considering the time available for a question to be solved in such an examination, is there a way to solve this problem without actually having to expand the the given relation by cross-multiplying the numerators and denominators or even finding the zeroes of the given equation.

It is practical to recall that $z\mapsto\frac{1-z}{1+z}$ is an involution. In particular $$\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$$ is the sum of the reciprocal of the roots of $p\left(\frac{1-x}{1+x}\right)=-\frac{x^3-x^2+7x+1}{(1+x)^3}$, which is also the sum of the reciprocal of the roots of $x^3-x^2+7x+1$. By Vieta's formulas, this is $\color{red}{-7}$ (option 3.).