n^{ 3} – n = n(n^{2} – 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

⇒ (n^{3} – 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

⇒ (n^{3} – 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

⇒ (n^{3} – 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, n^{3} - n is divisible by 2 and 3 both,

**therefore n**^{3} – n is divisible by 6.