Question 38

For any positive integer $n$ , prove that $n^3-n$ is divisible by 6.

Solution:

Let $a=n^3-n\Rightarrow a=n.({\mathrm{n2}}^{}-1)$
$\Rightarrow a=n.(n-1)(n+1)$ $\text{[}\therefore ({\mathrm{a2}}^{}-{\mathrm{b2}}^{})=(a-b\left)\right(a+b)\text{]}$…(i)
$\Rightarrow a=(n-1).n.(n+1)$
We know that,
I. If a number of completely divisible by 2 and 3, then it is also divisible by 6.
II. If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
III. If one of the factor of any number is an even number, then it is also divisible by 2.
$\therefore a=(n-1).n.(n+1)$ [From Eq. (i)]
Now, sum of 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 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 $n^3-n$ is always divisible by 6, where n is any positive integer.