n³ – n is divisible by 6, for each natural number n.

Given; P(n) = n³ – n is divisible by 6.

P(0) = 0³ – 0 = 0 ; is divisible by 6.

P(1) = 1³ – 1 = 0 ; is divisible by 6.

P(2) = 2³ – 2 = 6 ; is divisible by 6.

P(3) = 3³ – 3 = 24 ; is divisible by 6.

Let P(k) = k³ – k is divisible by 6.; ⇒ k³ – k = 6x.

⇒ P(k+1) = (k+1)³ – (k+1)

= (k+1)(k²+2k+1−1)

= k³ + 3k² + 2k

= 6x+3k(k+1) [n(n+1) is always even and divisible by 2]

= 6x + 3×2y ; is divisible by 6.

⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction P(n) = n³ – n is divisible by 6, for each natural number n.