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.

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.