Answer
$\left(n-\dfrac{1}{3}\right)\left(n^2+\dfrac{n}{3}+\dfrac{1}{9}\right)$
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, then,
\begin{array}{l}
n^3-\dfrac{1}{27}
\\\\=
\left[(n)+\left(-\dfrac{1}{3} \right) \right]\left[(n)^2-n\left(-\dfrac{1}{3} \right)+\left(-\dfrac{1}{3}\right) \right]
\\\\=
\left(n-\dfrac{1}{3}\right)\left(n^2+\dfrac{n}{3}+\dfrac{1}{9}\right)
.\end{array}
