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

Answer

N/a

Explanation

`a^(n)-b^(n)`
`={(a-b)+b}^(n)-b^(n)`
`=[.^(n)C_(0).(a-b)^(n)+^(n)C_(1).(a-b)^(n-1).b+^(n)C_(2)`
`.(a-b)^(n-2).b^(2)+.....+^(n)C_(n-1)(a-b)`
`.b^(n-1)+^(n)C_(n).b^(n)] -b^(n)`
`=(a-b)[.^(n)C_(0)(a-b)^(n-1)+^(n)C_(1).(a-b)^(n-2).b`
`+^(n)C_(2)(a-b)^(n-3).b^(2)+....+^(n)C_(n-1).b^(n-1)]+b^(n)-b^(n)`
`=(a-b)[.^(n)C_(0)(a-b)^(n-1)+^(n)C_(1).(a-b)^(n-2).b`
`+^(n)C_(2)(a-b)^(n-3).b^(2)+....+^(n)C_(n-1).b^(n-1)]`
which is divisible by (a-b)
So. `(a^(n)-b^(n))` is divisible by (a-b) for all positive intergers n.