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