I found this question from an old math olympiad questionnaire.

Let A, B, and C be the roots of $$x^3-4x-8=0$$ Find the numerical value of the expression $$\frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2}$$

I tried using Rational Roots Theorem. And then, I realized this equation involves complex numbers. Afterwhich, I tried various manipulations. But everytime I do manipulate I either go back from where I started or I went off too far and probably did something wrong in my calculations. I also tried tried doing something like "let u be x squared" and then tried manipulations there. I did found a value for x which are 0,0,+/-2 sqrt of 3, and I know these are wrong, so I gave up.

I do not know how to continue or what else to try. Can anyone help?

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    $\begingroup$ You need not find the exact value of each root. You should use Viete's theorem instead since the expression is symmetric. $\endgroup$ – Vim Aug 29 '16 at 8:42
  • $\begingroup$ Personally, I found your hint easier to grasp than Muralidharan and lab's solutions. I just think their works were too advanced for me. Thank you very much. $\endgroup$ – Carl Terence Valdellon Aug 29 '16 at 9:16
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    $\begingroup$ you're welcome. Well I'd say theirs are actually easier to follow because mine is just an idea but theirs span the idea into concrete steps. $\endgroup$ – Vim Aug 29 '16 at 9:20

\begin{align*} \frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2} &= \frac{A-2+4}{A-2}+\frac{B-2+4}{B-2}+\frac{C-2+4}{C-2} \\ &=3 +\frac{4}{A-2}+\frac{4}{B-2}+\frac{4}{C-2} \end{align*} The equation with $\alpha - 2, \beta -2, \gamma - 2$ as roots (where $\alpha, \beta, \gamma$ are the roots of the given equation) is $(y+2)^3 - 4(y+2) - 8 = y^3 + 6y^2 + 8y - 8 = 0$ and we want the sum of the reciprocals of the roots of this equation. This is $\frac{8}{8} = 1$. Thus \begin{align*} \frac{A+2}{A-2}+\frac{B+2}{B-2}+\frac{C+2}{C-2} &= 3+4 = 7 \end{align*}

  • $\begingroup$ Thank you for your help. Honestly, I do not quite understand your solution fully however I really do wish to understand it. May I know how you got the second equation from your first? I did not quite get that part. $\endgroup$ – Carl Terence Valdellon Aug 29 '16 at 9:18
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    $\begingroup$ Since $\alpha,\beta, \gamma$ satisfies the given equation, we need the equation satisfied by $\alpha-2, \beta-2, \gamma-2$. Put $y = x-2$. Then $x=y+2$. Substituting $y+2$ in the given equation, we obtain the equation satisfied by $\alpha-2, \beta-2, \gamma-2$. $\endgroup$ – user348749 Aug 29 '16 at 9:21
  • $\begingroup$ I did not understand the reason why you have to change (A+2/A-2)+(B+2/B-2)+(C+2/C-2) into 3+(4/A-2)+(4/B-2)+(4/C-2). I know how you did it. I just do not know why. Also, why should we get the sum of the reciprocals? Is it a general rule or am I just missing something out? $\endgroup$ – Carl Terence Valdellon Aug 29 '16 at 9:47
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    $\begingroup$ Given an equation, it is easier to compute the symmetric expressions involving roots using Vieta's formulas. In fact, if $\alpha,\beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$, then $\alpha+\beta+\gamma = -a, \alpha\beta+\beta\gamma+\gamma\alpha = b, \alpha\beta\gamma = -c$. You can see that $\frac{1}{\alpha} +\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma} = -\frac{b}{c}$. This is what we have used. $\endgroup$ – user348749 Aug 29 '16 at 9:54
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    $\begingroup$ You may see that we first formed the equation whose roots are $A-2, B-2, C-2$ and used the formula above to compute $\frac{1}{A-2}+\frac{1}{B-2}+\frac{1}{C-2}$ which is much easier than simplifying the given expression directly and applying Vieta's formulas. These "tricks" you learn as you go by. $\endgroup$ – user348749 Aug 29 '16 at 9:56

Let $y=\dfrac{x+2}{x-2}$

Using Componendo and Dividendo, $\dfrac{y+1}{y-1}=\dfrac x2\iff x=?$

Replace the value of $x$ in the given equation to form a cubic equation in $y$


Use Veita's formula to find the required sum $$=\dfrac71$$

  • $\begingroup$ Thank you for your help. I also do not understand your solution since I am not very familiar with the Componendo and Dividendo. But, I do really love to understand your solution. My first question: why did you set y as x+2/x-2 in the first place? $\endgroup$ – Carl Terence Valdellon Aug 29 '16 at 9:29
  • $\begingroup$ @CarlTerenceValdellon, For the last Question, observe what we need to calculate. Read the link on Componendo and Dividendo. $\endgroup$ – lab bhattacharjee Aug 29 '16 at 9:41
  • $\begingroup$ @labbhattarjee: Nice ref! (+1) $\endgroup$ – Markus Scheuer Aug 30 '16 at 7:20
  • $\begingroup$ @CarlTerenceValdellon, See also : iitjeemathematics.com/conte/quadratic-equations/… $\endgroup$ – lab bhattacharjee Aug 30 '16 at 7:58

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