Finding the roots of a 3rd degree polynomial 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? 

\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*}
