Question

For any +ve int. n, prove that n3(cube) - n is divisible by 6.

Answer 1

Hello friend

Here's your answer

(n³-n)=n(n²-1)

n²-1² is of the form a²-b²

So n³-n=n(n+1)(n-1)

We shall prove it by division algorithm

a=bq+r

When b=3, r can take the values 0,1,2

When r=0

a=3n(it is divisible by 3)
a-1=3n-1(it isn't divisible by 3)
a+1=3n+1(it isn't divisible by 3)

When r=1

a=3n+1(it isn't divisible by 3)
a-1=3n(it is divisible by 3)
a+1=3n+2(it isn't divisible by 3)

When r=2

a=3n+2(it isn't divisible by 3)
a-1=3n+1(it isn't divisible by 3)
a+1=3n+3(it is divisible by 3)

It is now clear that one among n,n-1,n+1 is divisible by 3

So n*(n-1)*(n+1) is divisible by 3 that is n³- n divisible by 3
_______________________

When b=2 r can take the values 0,1

When r=0

a=2q(it is divisible by 2)
a-1=2q-1(it isn't divisible by 2)
a+1=2q+1(it isn't divisible by 2)

When r=1

a=2q+1(it isn't divisible by 2)
a-1=2q(it is divisible by 2)
a+1=2q+2(it is divisible by 2)

It is now clear that one among n,n-1,n+1 is divisible by 2

So n*(n-1)*(n+1) is divisible by 2 that is n³-n is divisible by 2

As their product is divisible by both 2,3 it is also divisible by 6

So n³-n is divisible by 6

Answer 2

Please click on the file given below to see the solution.

Thank you!
(File will be added within a minute)