Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) 

We have n³ – n = n(n²-1)= (n-1)n(n+1), 

For n= 6q, 

⇒ (n-1)n(n+1)= (6q-1)(6q)(6q+1) 

⇒ (n-1)n(n+1)= 6[(6q-1)q(6q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q-1)q(6q+1)] 

For n= 6q+1, 

⇒ (n-1)n(n+1)= (6q)(6q+1)(6q+2) 

⇒ (n-1)n(n+1)= 6[q(6q+1)(6q+2)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)] 

For n= 6q+2, 

⇒ (n-1)n(n+1)= (6q+1)(6q+2)(6q+3) 

⇒ (n-1)n(n+1)= 6[(6q+1)(3q+1)(2q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)] 

For n= 6q+3, 

⇒ (n-1)n(n+1)= (6q+2)(6q+3)(6q+4) 

⇒ (n-1)n(n+1)= 6[(3q+1)(2q+1)(6q+4)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)] 

For n= 6q+4, 

⇒ (n-1)n(n+1)= (6q+3)(6q+4)(6q+5) 

⇒ (n-1)n(n+1)= 6[(2q+1)(3q+2)(6q+5)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)] 

For n= 6q+5, 

⇒ (n-1)n(n+1)= (6q+4)(6q+5)(6q+6) 

⇒ (n-1)n(n+1)= 6[(6q+4)(6q+5)(q+1)] 

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+4)(6q+5)(q+1)] 

Hence, for any positive integer n, n³ – n is divisible by 6.