Divisibility by n for 3- digit number We have 3 numbers:
$\; \overline{xyz}=100x+10y+z$, $\; \overline{yzx}$ and $\; \overline{zxy}$, all 3 divisible by the same number $n$.
Prove that the number $P=x^3+y^3+z^3-3xyz$ is also divisible by n.
$n \mid 100x+10y+z$
$n \mid 100y+10z+x$
$n \mid 100z+10x+y$
$n \mid 111x+111y+111z \Leftrightarrow n \mid 111(x+y+z)$ but since 111 is not a prime, we can't say for sure if $n \mid (x+y+z)$.
Also $P = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$.
So if I managed to prove that $n \mid (x+y+z)$ I think it would be sufficient, right?
Can you help me?

COMMENT.-We have the linear system $$\begin{pmatrix}100&10&1\\1&100&10\\10&1&100\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}na\\nb\\nc\end{pmatrix}$$ whose determinant is equal to
$\Delta=998001=3^6\cdot37^2$. Calculations give
$$x=\frac{n999(10a-b)}{3^6\cdot37^2}=\frac{n(10a-b)}{3^3\cdot37}(\text{ note that }3^3\cdot37=\sqrt{\Delta})$$ and
$$y=\frac{n(10b-c)}{3^3\cdot37}\\z=\frac{n(10c-a)}{3^3\cdot37}$$
Therefore we have after calculations $$x^3+y^3+z^3-3xyz=\frac{n^3}{\Delta}(a^3+b^3+c^3-3abc)$$ This shows that if the proposition is true then $$\frac{n^2}{\Delta}(a^3+b^3+c^3-3abc)\in \mathbb Z$$ This should not be difficult to prove while holding clearly from the system the equality $n(a+b+c)=3\cdot37(x+y+z)$.
