#### Answer

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

#### Work Step by Step

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}