Question

for any integer n, show that ncube -nis divisible by 6

Answer 1

We will prove this by mathematical induction..

Take an integer for example
$n = 1$
So the equations value is 0 which is divisible by 6...

So,
$p(1)$
is divisible by 6.

Let us assume for some integer x this is true, such that
$p(x) = 6k$
or

${x}^{3} - x = 6k$
where k is some integer..

So for integer x+1,
$p(x + 1) = ( {x + 1})^{3} - (x + 1)$
which simplifies to

$p(x + 1) = {x}^{3} + 3 {x}^{2} + 2x$
or
$p(x + 1) = x( {x}^{2} + 3 {x} + 2 )$

Here
${x}^{2} + 3{x} +2$ can be factorised to
$( x + 1 )(x + 2)$

So, the equation transforms to
$p(x + 1) = x( x + 1 )(x + 2)$

Let x be an even integer, so x,x+2 are even and x+1 is odd, or let x+1 be even, so x+2 and x are odd..
So at least one of them is divisble by 2.

Also, either of x,x+1,x+2 ( as they are 3 consecutive numbers ), is divisible by 3., can take either forms,
3n,3n+1,3n+2; or
3n-1,3n,3n+1; or
3n-2,3n-1,3n..
Long story short, either of the numbers is divisible by 3 and 1/2 of the numbers is/are divisible by 2. So the resulting product is always divisible by 6.
or
$p(x + 1) = 6( q )$