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

Explanation

To prove that `(a-b)` is a factor of `(a^(n) - b^(n))`, it must be proved that `a^(n) - b^(n) = k(a-b)`, where k is some natural number.
`a^(n) = (a-b + b)^(n)`
`=[(a-b)+b]^(b)`
`= .^(n)C_(0)(a-b)^(n) + .^(n)C_(1)(a-b)^(n-1) b + "…….." + .^(n)C_(n-1) (a-b) b^(n-1)+.^(n)C_(n)b^(n)`
`= (a-b)^(n) + .^(n)C_(1)(a-b)^(n-1)b+"....."+.^(n)C_(n-1)(a-b)b^(n-1)+b^(n)`
`rArr a^(n)-b^(n)=(a-b)[(a-b)^(n-1)+.^(n)C_(1)(a-b)^(n-2)b+"......."+.^(n)C_(n-1)b^(n-1)]`
`rArr a^(n)-b^(n) = k(a-b)`
where, `k = [(a-b)^(n-1)+.^(n)C_(1)(a-b)^(n-2)b+"........"+.^(n)C_(n-1)b^(n)-1]` is a natural number.
Thus, `(a-b)` is factor of `(a^(n) - b^(n))`, where n is a positive integer.