This is the basic and exam oriented question from real numbers chapter in which we have been asked to show that that n³-n is divisible by 6, where n is a positive integer.

Kindly solve the above problem

RS Aggarwal, Class 10, chapter 1A, question no 8

n³−n=n(n²−1)=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)=n³−n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)