Answer
$\dfrac{x^2+2x+4}{4}$
Work Step by Step
Factoring the expressions and then cancelling the common factors between the numerator and the denominator result to
\begin{array}{l}
\dfrac{x^3-8}{4x-8}
\\\\=
\dfrac{(x-2)(x^2+2x+4)}{4(x-2)}
\text{...cancel $(x-2)$}
\\\\=
\dfrac{x^2+2x+4}{4}
.\end{array}