For any positive Integer n, prove that ${{\text{n}}^3} - {\text{n}}$ is divisible by 6.

Hint – Simplify the given equation. Check for the divisibility of the equation with 3 and 2 as 6 = 2 × 3.

Complete step by step answer:
Simplify $n^3 – n$
$n^3 – n = n (n^2 - 1)$
            = 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.
⇒ ${\text{n}}\left( {{\text{n - 1}}} \right)\left( {{\text{n + 1}}} \right)$ 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.
⇒${\text{n}}\left( {{\text{n - 1}}} \right)\left( {{\text{n + 1}}} \right)$ is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴${\text{n}}\left( {{\text{n - 1}}} \right)\left( {{\text{n + 1}}} \right)$ = n3 - n is divisible by 6. (If a number is divisible by both 2 and 3, then it is divisible by 6)

Note: A number is divisible by 6 if it is divisible by both 2 and 3. I.e. if a number is divisible by two distinct numbers m and n, then that number should be divisible by m × n, which is a product of both numbers.