For any positive integer n , prove that n^3 − n divisible by 6.

Another solution  would be:

n^3 - n = n (n^2 - 1) = n (n - 1) (n + 1)

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
 
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n^3 - n is divisible by 6.( If a number is divisible by both 2 and 3, then it is divisible by 6)

n³ – n = n (n² – 1) = n (n – 1) (n + 1)

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q +
2 or, 6q + 3 or, 6q + 4 or, 6q + 5.
If n = 6q, then
(n − 1)(n)(n + 1) = (6q − 1)(6q)(6q + 1)
= 6[(6q − 1)(q)(6q + 1)]
= 6m, which is divisible by 6
If n = 6q + 1, then
(n − 1)(n + 1) = (6q)(6q + 1)(6q + 2)
= 6[(q)(6q + 1)(6q + 2)]
= 6m, which is divisible by 6
If n = 6q + 2, then
(n − 1)(n)(n + 1) = (6q + 1)(6q + 2)(6q + 3)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6
If n = 6q + 3, then
(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)
= 6[(3q + 1)(2q + 1)(6q + 4)]
= 6m, which is divisible by 6
If n = 6q + 4, then
(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)
= 6[(2q + 1)(3q + 2)(6q + 5)]
= 6m, which is divisible by 6
If n = 6q + 5, then
(n − 1)(n)(n + 1) = (6q + 4)(6q + 5)(6q + 6)
= 6[(6q + 4)(6q + 5)(q + 1)]
= 6m, which is divisible by 6
Hence, for any positive integer n, n³ – n is divisible by 6.​​