Proove that for every $\displaystyle $n \in \mathbb{N}$$
$\displaystyle n^3-9n+27\not\equiv 0 \mod 81$
Suppose otherwise, then there exists a $\displaystyle k\in \mathbb{Z}$ such that:
$\displaystyle n^3-9n+27=k\times 81=k\times 9^2$
which implies that $\displaystyle 9|n^3$ which in turn implies that $\displaystyle 3|n$.
Hence there exists a $\displaystyle \lambda \in \mathbb{N}$ such that:
$\displaystyle 27 \lambda^3-27 \lambda +27=3 \times k \times 27$
or:
$\displaystyle \lambda^3- \lambda +1=3 \times k$
Now consider the left hand side modulo 3.
CB