Question

for any positive integer n,prove that n3-n is divisible by 6

Answer 1

n3-n can also be written as n ( n2-1 )i.e. n(n-1)(n+1).Hence n3-n is a product of 3 consecutive natural no.sFrom the 3 no.s one of the no. is likely to be even. hence the no. n3-n is div by 2 -----------(1)Any 3 consecutive natural no. can be written in the form - 3x , 3x+1,3x+2 where x is a positive integer.This shows that from any 3 consecutive no. one of the no. is divisible by 3. Hence n3 - n has a factor 3 (as one of the 3 consecutive no.s is div by 3) ------------(2)Eq (1) (2) show that n3-n is divisible both by 3 as well as 2.So n3-n is also divisible by 6

Answer 2

n^3 – n = n(n^2 –1)

= n(n + 1)(n – 1)

= (n – 1) n(n + 1)

= Product of three consecutive positive integers  Now, we have to show that the product of  three consecutive positive integers is divisible  by 6.

We know that any positive integer a is of the  form 3q, 3q + 1 or 3q + 2 for some integer q.

Let a, a + 1, a + 2 be any three consecutive  integers.

If a = 3q

a(a + 1)(a + 2) = 3q(3q + 1)(3q + 2)

= 3q (2r)

= 6qr, which is divisible by 6.

If a = 3q + 1

a(a + 1)(a + 2) = (3q + 1)(3q + 2)(3q + 3)

= (2r) (3)(q + 1)

= 6r(q + 1) which is divisible by 6

If a = 3q + 2

a(a + 1)(a + 2) = (3q + 2)(3q + 3)(3q + 4)  = multiple of 6 for every  q = 6r (say) which is divisible by 6.

Therefore the product of three consecutive  integers is divisible by 6.