Prove by induction that the sum of the cubes of three consecutive natural numbers is divisible by `9.`

Explanation

Let `P(n)=n^3+(n+1)^3+(n+1)^3`, where `n in N`.
Step I For `n=1`.
`P(1)=1^3+2^(3)+3^3=1+8+27=35`, which is divisible by 9.
Step II Assume that P(n)is true for `n=k` , then
`P(k)=k^3+(k+1)^3+(k+2)^3`, where `k in N`.
`rArr P(k) =9r`, where r is a positive integer .
Step III For `n=k+1`.
`P(k+1)=(k+1)^3+(k+2)^3+(k+3)^3`
Now , `P(k+1)-P(k)=(k+1)^3+(k+2)^3+(k+3)^3-{k^3+(k+1)^3+(k+2)^3}`
`=(k+3)^3-k^3`
`=k^3+9k^2+27k+27-k^3`
`=9(k^2+3k+3)`
`rArr P(k+1)=P(k)+9(k^2+3k+3)`
`=9r+9(k^2+3k+3)`
`=9(r+k^2+3k+3)`
which is divisible by 9 as `(r+k^2+3k+3)` is a positive integer .
Hence , by the principle mathematical induction , P(n) is divisible by 9 for all `n in N`.