Properties of modular arithmetic clarification The sum of cubes of any three consecutive integers is divisible by $9$. i.e $(n^3 + (n+1)^3 + (n+2)^3)\mod9=0$ for every $n\in\mathbb{Z}$. 
They say it is sufficient to check out for values of $n=0,1,2,3,...8$ due to properties of modular arithmetic. Why this is so ?
Thanks

Alternatively:
$n^3+(n+1)^3+(n+2)^3$
$=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8$
$=3n^3+9n^2+15n+9$
$=9(n^2+1)+3n(n^2+5)$
In fact, you only need to check for $n=1;n=2;n=3$. 


*

*If $n=1$ or $n \equiv 1 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$

*If $n=2$ or $n \equiv 2 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$

*If $n=3$ or $3|n$ then $3|n^2+5\Rightarrow9|n(n^2+5)$
Note that $9|9(n^2+1)$, this finishes the proof. 
