**Question:**

**n3 – n is divisible by 6, for each natural number n ≥ 2.**

**Solution:**

According to the question,

P(n) = n3 – n is divisible by 6.

So, substituting different values for n, we get,

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

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

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

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

Let P(k) = k3 – k be divisible by 6.

So, we get,

⇒ k3 – k = 6x.

Now, we also get that,

⇒ 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) Which is divisible by 6, where y = k(k+1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

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