.
n^3+(n+1)^3+(n+2)^3
If n=1, :. n^3+(n+1)^3+(n+2)^3=1+8+27=36=4(9)
If n=2, :. n^3+(n+1)^3+(n+2)^3=8+27+64=99=11(9)
If n=3, :. n^3+(n+1)^3+(n+2)^3=27+64+125=216=24(9)
We have tried a few values for n and the expression has become divisible by 9.
Now we try n=k and based on our previous experience we will assume that the expression will be divisible by 9:
n=k, :. n^3+(n+1)^3+(n+2)^3=k^3+(k+1)^3+(k+2)^3=9m
k^3+k^3+3k^2+3k+1+k^3+6k^2+12k+8
k^3+(k+1)^3+(k+2)^3=3k^3+9k^2+15k+9=9m color(red)(Equation-1)
Now we will try n=k+1. If we can prove that in this case the expression is still divisible by 9 then we can declare that the original expression is divisible by 9 for all values of n.
(k+1)^3+(k+2)^3+(k+3)^3=k^3+3k^2+3k+1+k^3+6k^2+12k+8+k^3+9k^2+27k+27=3k^3+18k^2+42k+36
We can break up the expression as such:
(k+1)^3+(k+2)^3+(k+3)^3=color(red)(3k^3+9k^2+15k+9)+9k^2+27k+27
As evident, the first four terms that are shown in red are equal to 9m according to color(red)(Equation-1).
Therefore,
(k+1)^3+(k+2)^3+(k+3)^3=9m+9(k^2+3k+3)
(k+1)^3+(k+2)^3+(k+3)^3=(m+k^2+3k+3)(9)
which shows it to be divisible by 9 and the problem is solved.