Question

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

Answer 1

Let a = n³ – n

`\implies` a = n – (n² – 1)

`\implies` a = n – (n – 1)(n + 1)  ......[∵ (a² – b²) = (a – b)(a + b)]

`\implies` a = (n – 1) n (n + 1)

We know that, if a number is divisible by both 2 and 3, then it is also divisible by 6.

n – 1, n and n + 1 are three consecutive integers.

Now, a = (n – 1 ) n (n + 1) is product of three consecutive integers.

So, one out of these must be divisible by 2 and another one must be divisible by 3.

Therefore, a is divisible by both 2 and 3.

Thus, a = n³ – n is divisible by 6.