Let `K = n(n^2-1)`

For n=1;

`K = 1(1^2-1) = 0 = 6*0`

So when n=1, K is divisible by 6.

Let us assume for a positive number n=p the expression K is divisible by 6.

`K = p(p^2-1) = 6*r` where r is a positive integer.

For n=p+1

`K = (p+1)[(p+1)^2-1]`

`= (p+1)(p^2+2p)`

`= p^3+3p^2+2p`

`= p^3+3p^2+2p+p-p`

`= p^3-p+3p^2+3p`

`= p(p^2-1)+3p(p+1)`

If p is any positive number (p+1) is also positive. If p is odd then (p+1) is even. If p is even then (p+1) is odd. The product of odd number and a even number is a even number.

for eg: 2*3 = 6

4*5 = 20

So p(p+1) is a even number. We can write p(p+1) = 2*q where q is a positive integer.

`K = p(p^2-1)+3p(p+1)`

`= 6*r+3*2*q`

`= 6*r+6*q`

`= 6(r+q)`

So when n=p+1 ; K is divisible by 6.

**When n=1 ; K is divisible by 6**

**When we assume for positive n=p ; K is divisible by 6 then for n=p+1 it becomes true.**

**So from mathematical induction for all positive ` n in Z` the result is true.**

Notice that you may write the difference of squares `n^2 - 1` as a product such that:

`n^2 - 1 = (n -1)(n + 1)`

Hence, the given product `n(n^2 - 1) ` may be considered as a product of three consecutive numbers such that:

`n(n^2 - 1) = (n -1)n(n + 1)`

Hence, for `n = 1 =gt 0*1*2 = 0 =gt0` is divisible by 6

Hence, for `n =2 =gt 1*2*3 =6` `=gt6` is divisible by 6

Notice that the first step of the technique called mathematical induction is checked, hence, you may consider the second step called the inductive step:

If P(k) is true, then you need to prove that P(k+1) is true.

`P(k) = (k-1)k(k+1)` considered true => `(k-1)k(k+1) = 6p =gt k(k+1) = 6p/(k-1)` `P(k+1) = (k+1-1)(k+1)(k+1+1) =gt P(k+1) = k(k+1)(k+2)`

You need to substitute `6p/(k-1)` for `k(k+1), ` such that:

`P(k+1) = (6p/(k-1))(k+2)`

**It is clear now that `P(k+1)= (6p/(k-1))(k+2)` is divisible by, hence `P(n) = n(n^2 - 1)` is divisible by 6 for any integer positive value of n.**

When n=1;

n(n^2-1)=(1)((1)^2-1)=0; which can be divided by 6.

Therefore the result is OK for when n=1.

Assume that the result is OK for n=P;

Therefore;

p(p^2-1)=6k.........(1) (k is an positive intiger)

When n=p+1,

(p+1)[(p+1)^2-1]=p(p+1)^2-p+(p+1)^2-1

=p(p^2+2p+1)-p+(p^2+2p+1)-1

=P(p^2+2p+2-1)-p+p^2+2p+1-1

=P(p^2-1)+p(2p+2)+p^2+p

=P(p^2-1)+2p^2+2p+p^2+p

=p(p^2-1)+3p^2+3p

=p(p^2-1)+3p(p^2+1)

=p(p^2-1)+3p(P^2-1+2)

=p(p^2-1)+3p(P^2-1)+6p

=4p(P^2-1)+6p

=4*6k+6p ; Where p(P^2-1)=6k (from (1))

=6(4k+p) ; can be divided by 6.

Therefore the result is OK for n=p+1,

since the result is OK for n=1, n=p and n=p+1 from the mathematican induction the result is OK for any positive value of n.