for any positive integer n,p prove that n^3-n is divisible by 6
Answers
Answered by
2
n^3 - n = n(n^2-1)
= (n-1)n(n+1)
These are 3 consecutive integers
So one of them must be divisible by 3
Similarly , one/two of them must be divisible by 2
Thus , the whole term becomes divisible by 3*2=6
(There are theorems for the same but rather focus on quick answers than applying them)
= (n-1)n(n+1)
These are 3 consecutive integers
So one of them must be divisible by 3
Similarly , one/two of them must be divisible by 2
Thus , the whole term becomes divisible by 3*2=6
(There are theorems for the same but rather focus on quick answers than applying them)
Similar questions
India Languages,
8 months ago
Math,
8 months ago
Physics,
8 months ago
Computer Science,
1 year ago
Chemistry,
1 year ago
History,
1 year ago