## Precalculus (6th Edition) Blitzer

$(x+4)(x^2-4x+16)$
The given expression can be written as $x^3+4^3.$ RECALL: $a^3+b^3=(a+b)(a^2-ab+b^2).$ Factor the given sum of two cubes to obtain $(x+4)(x^2-x(4)+4^2) \\=(x+4)(x^2-4x+16).$