Finding the roots of a 3rd degree polynomial

\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*}

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 $$=\dfrac71$$