Prove that the sum of cubes of three consecutive natural numbers is always divisible by 3.

Given: We have information that, we can assume three consecutive natural numbers.

To Prove: Sum of the cubes of three consecutive natural numbers is always divisible by 3.

Proof: Let three consecutive numbers be x, (x + 1) and (x + 2).

Sum of cubes of these numbers = x³ + (x + 1)³ + (x + 2)³

⇒ Sum of cubes of these numbers = x³ + x³ + 1 + 3x (x + 1) + x³ + 2³ + 3x (x + 2)

⇒ Sum of cubes of these numbers = x³ + x³ + 1 + 3x² + 3x + x³ + 8 + 6x² + 12x

⇒ Sum of cubes of these numbers = 3x³ + 9x² + 15x + 9

⇒ Sum of cubes of these numbers = 3 (x³ + 3x² + 5x + 3)

This clearly shows that the expression, 3 (x³ + 3x² + 5x + 3) is divisible by 3.

Thus, proved that the sum of cubes of three consecutive natural numbers is always divisible by 3.