Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$ 
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$

The concept of transformation of roots can be applied here.
So replace
$$x \to \frac{x-1}{x+1}$$
After considerable algebra, a cubic would be obtained, and then the sum can easily be found out
My question: This method is touted around a lot, but I never around how it actually works, especially in this case. If $\frac{x-1}{x+1} =\alpha$, then the transformed equation would be $\alpha^3 -2007 \alpha +2002$, and the sum of all $\alpha$ should still be $2007$. Can I get an explanation for this method?

If $x$ is any of $a,b,c$ then plugging in $x=\frac{1+y}{1-y}$ gives the new equation $$y^3 - 2y^2 + \frac{999}{1002} \,y+\frac{1}{1002}=0 \, .$$ Finding the objective then amounts to finding the sum $A+B+C$ for the 3 solutions of this equation which is just the negative of the coefficient of the $y^2$ term in the general cubic $$(y-A)(y-B)(y-C)=y^3-(A+B+C)y^2 + (AB+AC+BC)y - ABC=0 \, ,$$ i.e. $A+B+C=2$.
