The sum of the cubes of three consecutive natural numbers is divisible by

Question

Aicharya · Accepted Answer

Let the three consecutive natural numbers $(n-1), n, n+1$ where $n \geq 2$ $\therefore$ Sum of cubes of the three consecutive natural numbers is $(n-1)^3+n^3+(n+1)^3=3 n^3+6 n=3 n\left(n^2+2\right)$ For each value of $n \geq 2$, the expression $3 n\left(n^2+2\right)$ is divisible by 9 . Hence, option (c) is correct.

The sum of the cubes of three consecutive natural numbers is divisible by

Solution