If a and b are distinct integers then prove that (a-b) is a factor of `(a^(n)-b^(n))`, whenever n is a positive integar.

Explanation

We may write `a^(n) = ( a-b+b)^(n) = {(a-b) + b^(n)}`
`=^(n)C_(0) xx (a-b)^(n) + .^(n) C _(1) xx ( a - b)^((n-1)) xx b + .^(n)C_(2) xx (a-b)^((n-2)) xx b^(2) + `
`...+ .^(n)C_(n-1) xx (a-b) xx b^(n-1) + .^(n) C_(n) xx b^(n)`
`=(a-b)^(n) + nb(a-b)^(n-1)+ 1/2 n (n-1)b^(2) (a-b)^(n-2) +`
`... + nb^(n-1) (a-b) + b^(n)`
`rArr (a^(n) - b^(n)) = (a-b)^(n) + nb(a-b)^(n-1) + 1/2 n ( n-1)b^(2) (a-b)^(n-2) + `
`... + nb^(n) (a-b)`
`=(a-b) xx [(a - b)^(n-1) + nb ( a-b) ^(n-2)`
`=1/2 n (n-1) b^(2) ( a-b)^(n-3) + nb^(n)]`
[taking `(a-b)` common throughout]
`=(a-b) xx "(an integar)"`
Hence, `(a-b) " is a factor of " (a^(n) - b^(n)).`