Note that the residue of $n^3$ modulus $9$ cycles between $0, 1,$ and $-1$ as $n$ increases by $1$. Thus $n^3 + (n+1)^3+(n+2)^3 \equiv 0+1+(-1) \equiv 0\pmod{9}$.

batter8642 wrote: Show that for any natural number n, n^3 + (n + 1)^3 + (n + 2)^3 is divisible by 9. $n^3+(n+1)^3+(n+2)^3=(n+n+1+n+2)(n^2+(n+1)^2+(n+2)^2-n(n+1)-(n+1)(n+2)-(n+2)n)+3n(n+1)(n+2)=9(n+1)+3n(n+1)(n+2)$.Of course now it is divisible by $9$ and also if $n$ is odd, it's divisible by $18$.

use induction

batter8642 wrote: Show that for any natural number n, n^3 + (n + 1)^3 + (n + 2)^3 is divisible by 9. there could be three cases.namely $n=3k ;3k+1 ;3k+2$.So now if$n=3k \implies n^3+(n+1)^3+(n+2)^3\equiv 0+1+(-1)=0 \mod(9) ;n=3k+1 \implies n^3+(n+1)^3+(n+2)^3\equiv 1+(-1)+0=0 \mod(9) ;n=3k+2 \implies n^3+(n+1)^3+(n+2)^3\equiv (-1)+0+1=0 \mod(9)$ and hence done.