\(\dfrac{x^3+64}{2x^3-8x^2+32x}\\ =\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\\ =\dfrac{x+4}{2x}\)
\(\dfrac{x^3+64}{2x^3-8x^3+32x}\)
\(=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\)
\(=\dfrac{x+4}{2x}\)