Finding the roots of a 3rd degree polynomial

Question

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?

Answer 1

\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*}

Answer 2

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 $y$

$$(2-1)y^3-(4+3)y^2+\cdots=0$$

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