**Question:**

For any positive integer n, prove that n3 – n is divisible by 6.

**Solution:**

Let $a=n^{3}-n \Rightarrow a=n \cdot\left(n^{2}-1\right)$

$\Rightarrow \quad a=n \cdot(n-1) \cdot(n+1)$ $\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]$

$\Rightarrow \quad a=(n-1) \cdot n \cdot(n+1)$ $\ldots$ (i)

We know that,

- If a number is completely divisible by 2 and 3, then it is also divisible by 6.
- If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
- If one of the factor of any number is an even number, then it is also divisible by 2.

∴ a = (n – 1) . n . (n + 1) [from Eq. (i)]

Now, sum of the digits =n-1+n+n+1=3n

= multiple of 3, where n is any positive integer,

and (n -1)- n- (n + 1) will always be even, as one out of (n -1) or n or (n + 1) must of even. Since, conditions II and III is completely satisfy the Eq. (i).

Hence, by condition I the number n3 – n is always divisible by 6, where n is any positive

integer. Hence proved.