## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(x +3)(x^2 -3x +9)$
Since $x^3+27$ is a sum of the two cubes $x^3$ and $3^3$, then by making use of the sum of two cubes $$a^3 + b^3 = (a + b)(a^2 - ab + b^2),$$ we have \begin{align*} x^3 +27&=x^3+3^3\\ &= (x +3)(x^2 -3x + 3^2)\\ &= (x +3)(x^2 -3x +9) \end{align*}