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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?

Aditya
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    I'd be tempted to do a bit of simplification before making that substitution. Noticing that $\frac{a-1}{a+1} = 1 - \frac{2}{a+1}$, I'd hope to rewrite the cubic using $x \to \frac{1}{x+1}$. That would be combining the shift $x \to x+1$ and the root reciprocals map (which is just reversing the order of coefficients). – hardmath Aug 22 '20 at 15:44
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    In any case just what part of the method needs explanation? If your last transformed equation were correct (I haven't checked), the sum of roots would be zero, not $2007$. – hardmath Aug 22 '20 at 15:46
  • When we transform, we really just take the form of the root that we want into a new equation. For example, if you were given the equation $2^{2x}+2^x=\pi$. What would you do? First thought would be to substitute for $2^x=y$ and then solve the quadratic in $y$. You are doing the same thing here, just in a more confusing and convoluted way. – Popular Power Aug 22 '20 at 16:06
  • @PopularPower what would be the more efficient method? – Aditya Aug 22 '20 at 16:34

4 Answers4

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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 $a+1,b+1,c+1$.

hardmath
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  • Would the substitution I used work as well? – Aditya Aug 30 '20 at 18:00
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    Aditya: It's easy to get "turned around" in this substitution method. You proposed that replacing $x$ by $(x-1)/(x+1)$ we'd get an equation whose roots are $$(a-1)/(a+1),(b-1)/(b+1),(c-1)/(c+1)$$ But this is backwards. When we replace $x$ with $f(x)$, the roots of the resulting equation are values that satisfy $f(x) = $ root of the original equation. So what you'd need is the functional inverse of $y=(x-1)/(x+1)$, which is $x=(1+y)/(1-y)$. This is how @Diger tackles the problem (and gets the right answer). – hardmath Aug 30 '20 at 18:19
  • I am having trouble understand your point, on why we need a functional inverse. Can you elaborate on that? (Namely ‘the roots of the resulting equation are values that satisfy () root of the original equation‘ – Aditya Aug 30 '20 at 18:28
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    If $p(x) = 0$ and $y = f(x)$, then $p(f^{-1}(y)) = 0$. I used a simple case of this when I made the substitution of $x = y-1$ to get an equation at the beginning for $a+1,b+1,c+1$. – hardmath Aug 31 '20 at 04:07
  • So the zeroes obtain from $p(f^{-1})=0$ will be the zeroes for $f(x)$? – Aditya Aug 31 '20 at 07:48
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    No, that's not what I'm saying. Rather the implication is that if $x$ is a root of $p$, then $f(x)$ is a root of $p(f^{-1})$ (assuming of course that $f$ has a suitable function inverse). – hardmath Aug 31 '20 at 12:59
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    So basically the algorithm is to apply $y=$ whatever the desired function, and than write $x=f^{-1} (y)$ and then substituting $x$ in the given equation. Is that correct? – Aditya Aug 31 '20 at 17:55
  • I'll add an example to the Answer, to make sure we are seeing eye to eye. – hardmath Aug 31 '20 at 18:05
  • As it turns out, there was a sign error in sirswagger21's final value for the quotient. I think everyone now agrees on $ \ +2 \ \ . $ –  Dec 19 '21 at 05:07
  • Can you give an example where subsituting method will not work because the inverse doesnt exist ? And how to solve it @hardmath – Orion_Pax Apr 06 '22 at 04:09
  • @Orion_Pax: I can think of some simple cases, e.g. a constant function or $f(x) = x^2$. Would it not be better to ask a new Question that highlights the function of roots you are interested in? You can link this Question for context. – hardmath Apr 06 '22 at 06:07
  • Sir you mean whose roots are getting squared? But that can be solved by considering the √y = x substitution isnt ? And yeah it was just a thought so asked here next time will try asking as a new question – Orion_Pax Apr 06 '22 at 08:27
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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. $A+B+C=2$.

Diger
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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 $$\boxed{+2.}$$

Calvin Khor
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sirswagger21
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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

user600016
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