## Precalculus (6th Edition) Blitzer

$(x+3)(x^2-3x+9)$
The given expression can be written as $x^3+3^3.$ RECALL: $a^3+b^3=(a+b)(a^2-ab+b^2).$ Factor the given sum of two cubes using the formula above with $a=x$ and $b=3$ to obtain $(x+3)(x^2-x(3)+3^2) \\=(x+3)(x^2-3x+9).$