Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.2 - Factoring the Difference of Two Squares - Problem Set 6.2 - Page 249: 86

$(n-3)(n^2+3n+9)$

Using $a^3-b^3=(a-b)(a^2+ab+b^2)$, or the factoring of the difference of $2$ cubes, the factored form of the given expression, $ n^3-27 ,$ is \begin{array}{l}\require{cancel} (n-3)[(n)^2+n(3)+(3)^2] \\\\= (n-3)(n^2+3n+9) .\end{array}