<s>[INST] 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?
 [/INST] The equation is expanded to $$\frac{3abc+(ab+ac+bc)-(a+b+c)-3}{abc+(ab+ac+bc)+(a+b+c)+1}.$$
By Vieta's, $$abc = -2002$$ $$ab+ac+bc = -2007$$ $$a+b+c = 0$$
Therefore, the expression is $$\frac{3(-2002) + (-2007) - 0 - 3}{-2002 + (-2007) + 0 + 1}.$$
This simplifies to $$\boxed{+2.}$$
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