## Intermediate Algebra (12th Edition)

$(x+4)(x^2-4x+16)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $x^3+64 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using $(a\pm b)(a^2\mp ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} (x+4)[(x)^2-x(4)+(4)^2] \\\\= (x+4)(x^2-4x+16) .\end{array}