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?
 A: Since the two posted Answers so far (that provide a concrete value of the indicated sum) disagree, I'll pursue the approach I'd suggested in a Comment.
First form an equation whose three roots are $y = (x+1)$ instead of $x$.  That is:
$$ (y-1)^3 - 2007(y-1) + 2002 = 0 $$
$$ y^3 - 3y^2 - 2004y + 4008 = 0 $$
An equation whose three roots are instead $z = 1/y = 1/(x+1)$ is then:
$$ 4008z^3 - 2004z^2 - 3z + 1 = 0 $$
This is obtained by reversing the order of coefficients.  By inspection we have the sum of those three roots:
$$ z_1+z_2+z_3 = \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{2004}{4008} = \frac{1}{2} $$
Finally:
$$ \frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1} = 3 - 2\left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right) = 2 $$

It may seem counterintuitive that, in order to transform the roots of a polynomial equation $p(x) = 0$, we make a substitution using the functional inverse of that transform.  Let's sketch why this is so, and then do a simple example.
If we wanted instead of the original roots of $p(x)$, some function $f$ applied to each of those roots, consider what equation these transformed roots must satisfy.  That is, where $x$ satisfies $p(x) = 0$, what does the transformed value $\alpha = f(x)$ satisfy?  Clearly if the functional inverse $f^{-1}$ exists, we can say:
$$ p(f^{-1}(f(x))) = 0 $$
In other words the transformed value $\alpha$ satisfies:
$$ p(f^{-1}(\alpha) = 0 $$
Depending on how complicated $f^{-1}$ is, the substitution may cause us to do some careful algebra to simplify the new equation.  So to illustrate, let's do an easy example.
Define the polynomial $p(x) := x^2 - 2x$, which has roots for $p(x) = 0$ at $x = 0,2$.
If we wanted to transform those roots by adding one, then we need to substitute into $p(x)$ the opposite transform, subtracting one.  Then:
$$ p(x-1) = (x-1)^2 - 2(x-1) = x^2 - 4x + 3 = (x-1) (x-3) $$
Now we have roots of $p(x-1)$ at $x = 1,3$.  In other words we have transformed the roots of $p(x)$ by adding one to each of them, as we wished.
Note that this was essentially the first substitution I used in my solution, to get an equation with roots $a+1,b+1,c+1$.
A: 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. $A+B+C=2$.
A: 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 $$\boxed{+2.}$$
A: In general if you need to find $\sum_{cyclic} f(a)$, where f(a) is a function of a root, then the transformation that needs to be applied is $x \rightarrow f^{-1}(x)$. In the original polynomial equation.
The proof actually follows from definition of inverse of a function.
Note: This works only for polynomials. This can be extended to any cyclic sum/product/combination of the two of the roots of the polynomial equation
