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?
Best Answer
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 $a+1,b+1,c+1$.