Find the numerical value.

Given equation: $x^3-4x-8=0$ , where the roots are $a, b,$ and $c$

$a+b+c=0$

$ab+bc+ac=-4$

$abc=8$

The expression

$\frac{a+2}{a-2}+\frac{b+2}{b-2}+\frac{c+2}{c-2}$

$=\frac{a-2+4}{a-2}+\frac{b-2+4}{b-2}+\frac{c-2+4}{c-2}$

$=3+4(\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2})$

$=3+4(\frac{ac+bc+ab-4(a+b+c)+12}{abc-2(ab+bc+ac)+4(a+b+c)-8})$

$=3+4(\frac{-4-4(0)+12}{8-2(-4)+4(0)-8})$

$=3+4(\frac{8}{8})$

$=3+4$

$=\boxed{7}$

I know, direct application of vieta's relations work, but let's try a different approach

Given equation: $x^{3}-4x-8$ where the roots are $\text{a,b, and c}$

we need to find the value of $\sum\frac{x_k+2}{x_k-2}$ where $x_k$ are the roots of the given equation, i.e $x_k=a,b,c$

let $y_k=\frac{x_k+2}{x_k-2}$

therefore, by componendo dividendo,

$x_k=2\frac{y_k+1}{y_k-1}$

(the motivation for doing this was trying to find a polynomial with roots= $y_k$ and then calculate $\sum y_k$ by vieta's relation)

putting the values of $x_k$ in the given polynomial yields the equation:

$(\frac{y_k+1}{y_k-1})^{3} - (\frac{y_k+1}{y_k-1}) -1=0$

A short computation yields this as:

$y_k^{3}-7y_k^{2}-y_k-2=0$

By vieta's relations,

the desired sum $=\sum\frac{x_k+2}{x_k-2}=\sum y_k=7$

Our answer= $\boxed{7}$