n 3 – n = n(n2 – 1) = n(n – 1)(n + 1)
For a number to be divisible by 6, it should be divisible by 2 and 3 both,
Divisibility by 3:
n – 1, n and n + 1 are three consecutive whole numbers.
By Euclid’s division lemma
n + 1 = 3q + r, for some integer k and r < 3
As,
r < 3 possible values of ‘r’ are 0, 1 and 2.
If r = 0
n + 1 = 3q
⇒ n + 1 is divisible by 3
⇒ n(n – 1)(n + 1) is divisible by 3
⇒ (n3 – n) is divisible by 3
If r = 1
⇒ n + 1 = 3q + 1
⇒ n = 3q
⇒ n is divisible by 3
⇒ n(n – 1)(n + 1) is divisible by 3
⇒ (n3 – n) is divisible by 3
If r = 2
⇒ n + 1 = 3q + 2
⇒ n + 1 – 2 = 3q
⇒ n – 1 = 3q
⇒ n - 1 is divisible by 3
⇒ n(n – 1)(n + 1) is divisible by 3
⇒ (n3 – n) is divisible by 3
Divisibility by 2:
If n is even
Clearly,
n(n – 1)(n + 1) is divisible by 2
If n is odd
⇒ n + 1 is even
⇒ n + 1 is divisible by 2
⇒ n(n – 1)(n + 1) is divisible by 2
Hence,
For any value of n, n3 - n is divisible by 2 and 3 both,
therefore n3 – n is divisible by 6.