.
#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.
