Question

Prove that for all whole values of n the value of the expression (n-3)(n+2)-(n-3)(n+8) is divisible by 6.

Answer 1

Hence, n 3 −n=n(n+1)(n−1) is divisible by 6.

Explanation:

Let's simplify the equation first:

(n-3)(n+2)-(n-3)(n+8)

= n² - n - 6 - (n² + 5n - 24)

= n² - n - 6 - n² - 5n +24

= -6n - 18

Divisible means that the equation can be divided by 6 with no remainder.

If I divide the equation by 6, I get (-n-3)

It goes in evenly, therefore it is divisible by 6