Prove that induction that the sum of the cubes of three consecutive numbers is divisible by g.

Solution: To show that $(n-1)^3 + n^3 (n+1)^3$ is divisible by 9.

Let p(n) = $(n-1)^3 n^3 + (n+1)^3$ be the given statement.

Step 1. Verification step.

put n = 1 in p(n)

$\therefore p(1) = 0^3 + 1^3 + 2^3 = 9$

$\therefore p(1) = 9 $ is divisible by 9.

step 2. Inductive hypothesis.

put n = k in p(n)

$\therefore p(k) = (k-1)^3 + k^3 + (k+1)^3$ is assume to be divisible by 9 by using inductive hypothesis only.

Step 3. conclusion step.

put n = k +1 in p(n)

$\therefore p(k+1) = k^3 + (k+1)^3 +(k+2)^3$

$= k^3 + k^3 + 3k^2 + 3k+1 k^3 + 6k^2 + 12k + 8$

$\therefore p(k+1) = sk^3 + 3k^2 + 1 + 3k + 6k^2 + 12k + 8$

$= 3k^3 + 9k^2 + 15k + 9$

$= (3k^3 + 6k) (1) + 9 (k^2 + k +1) (2)$

Here term 1 is div by 9 by equation 1 and 2nd term is multiple of 9.

$\therefore$ p(k+1) is also divisible by 9

$\therefore p(n) = (n - 1)^3 + n^3 + (n+1)^3$ is divisible by 9. for all values of natural number.