Is your question to prove via mathematical induction that n3-n is divisible by 6 for all integers n?
If so, we shall first prove the above for the non-negative integers. First, we begin with a base case; let us take that as n=0. n=0 clearly satisfies the desired property since 03-0=0, which is divisible by 6.
Secondly, let us appeal to the inductive hypothesis: assume that for an integer n, n3-n divisible by 6. Then we wish to show the integer immediately after n also satisfies the property: (n+1)3-(n+1) is divisible by 6.
The above can be shown by expanding the above expression: (n+1)3-(n+1) = n3+3n2+3n+1-n-1= (n3-n)+(3n2+3n)=(n3-n)+3n(n+1). n3-n is divisible by 6 by assumption. Also, it is clear that 3n(n+1) is divisible by 6.
The second claim can be seen by the fact that since n is an integer, either n or n+1 will be odd (and correspondingly either n+1 or n will be even). Thus, representing the even number as 2N for some integer N, n(n+1) can also be written as 2N*(2N+1) (if n is even) or 2N*(2N-1) (if n is odd). Thus 3n(n+1)=3*2N*(2N+ or - 1)=6N(2N+ or - 1), which is clearly divisible by 6 [6N(2N + or - 1)/6=N(2N+ or - 1)].
Since the sum of two numbers are divisible by 6, the sum itself will be divisible by 6, making (n+1)3-(n+1) divisible by 6.
Having shown the seed case, and having shown that an integer satisfying the above property implies that the integer after that also satisfies the above property, we have proven the claim for all positive integers.
Now, assuming n is positive then -n is negative. Insert -n into the expression: (-n)3-(-n)=-n3+n=-(n3-n), which is simply the negative of the expression already dealt with before. Having proven n3-n is divisible by 6, its negative must also be divisible by 6. This proves the claim for all negative integers.