Answer
$x^2-4x+16$
$x\ne-4$
Work Step by Step
Given.
$\frac{x^3+64}{x+4}$
Write 64 as $4^3$.
$=\frac{x^3+4^3}{x+4}$
Factor the sum of two cubes using the formula $a^3+b^3=(a+b)(a-ab+b^2)$.
$=\frac{(x+4)(x^2-4x+16)}{x+4}$
Cancel the common factor $x+4$.
$x^2-4x+16,x\ne-4$
