Question

for any positive integer n prove that n cube minus n is divisible by 6​

Answer 1

n^3 - n

n( n^2 -1)

n( n-1) ( n+1)

now n,n-1,n+1 are three consecutive integers

so there would be a factor 3 always

As they are 3 consecutive So there would be factor 2 always

So it would be divisible by 6 always

or

n( n-1)( n+1)

now if n is odd

so n-1 and n+1 should be even

if both n-1 and n+1 is even then there would be factor 4

And they are 3 consecutive So 3 would br factor

So 12 would be factor which is divisible by 6

if n is even

n-1 and n+1 odd

n is even so 2 as factor

and 3 consecutive So 3 would be there as factor

So 6 as factor

So its divisible by 6