Answer
$\frac{8}{x(x-2)(x+2)}=\frac{-2}{x}+\frac{1}{x-2}+\frac{1}{x+2}$
Work Step by Step
$\frac{8}{x^3-4x}$, factorising $x^3-4x=x(x^2-4)=(x)(x-2)(x+2)$,
thus,
$\frac{8}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2}$,
$A(x-2)(x+2)+Bx(x+2)+Cx(x-2)$,
$A(x^2-4)+B(x^2+2x)+C(x^2-2x)$,
$(A+B+C)x^2+(2B-2C)x+(-4A)=8$,
$\begin{cases}
A+B+C=0\\
2B-2C=0\\
-4A=8
\end{cases}$,
thus, $A=-2$,
Multiplying Equation 1 by 2 and adding it to Equation 2.
$\begin{cases}
2B+2C=4\\
2B-2C=0\\
-- -- -- --\\
4B+0C=4
\end{cases}$
thus, $B=2=1$, substituting back into the Equation, $2-2C=0, C=1$.
Therefore,
$\frac{8}{x(x-2)(x+2)}=\frac{-2}{x}+\frac{1}{x-2}+\frac{1}{x+2}$
