Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.2 - Factoring the Difference of Two Squares - Problem Set 6.2 - Page 249: 87

$(n+4)(n^2-4n+16)$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum of $2$ cubes, the factored form of the given expression, $ n^3+64 ,$ is \begin{array}{l}\require{cancel} (n+4)[(n)^2-n(4)+(4)^2] \\\\= (n+4)(n^2-4n+16) .\end{array}