Algebra: A Combined Approach (4th Edition)

$\dfrac{x^{3}-8}{4x-8}=\dfrac{x^{2}+2x+4}{4}$
$\dfrac{x^{3}-8}{4x-8}$ Factor the numerator and take out common factor $4$ from the denominator and simplify: $\dfrac{x^{3}-8}{4x-8}=\dfrac{(x-2)(x^{2}+2x+4)}{4(x-2)}=\dfrac{x^{2}+2x+4}{4}$