#### Answer

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

#### Work Step by Step

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}