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Mathematics Question on Binomial Theorem for Positive Integral Indices

If a and b are distinct integers, prove that a - b is a factor of an−bna^n - b^n , whenever n is a positive integer.
[Hint: writean=(a−b+b)n a ^n = (a - b + b)^n and expand]

Answer

In order to prove that (a-b) is a factor of (an−bn)(a^n - b^n), it has to be proved that an−bn=k(an−bn)a^n - b^n= k (a^n - b^n), where k is some natural number
It can be written that, a= a - b + b
∴an(a−b+b)n=[(a−b)+b]n∴ a^n (a-b+b)^n =[(a-b)+b]^n
=C0n(a−b)n+C1n(a−b)n−1b+...+Cn−1n(a−b)bn−1+Cnnbn= C_0^n (a-b)^n +C_1^n (a-b)^{n-1} b+...+ C_{n-1}^n(a-b)b^{n-1} + C_n^n b^n
=(a−b)n+c1n(a−b)n−1b+...+Cn−1n(a−b)bn−1+bn=(a-b)^n +c_1^n (a-b)^{n-1} b+...+ C_{n-1}^n (a - b) b^{n-1}+b^n
⇒an−bn=(a−b)[(a−b)n−1+C1n(a−b)n−2b+...+Cn−1n bn−1]⇒ a^n − b^n = (a−b)[(a−b)^{n-1} + C_1^n (a - b)^{n-2} b + ...+ C_{n-1}^n\,b^{n-1}]
⇒an−bn=k(a−b)⇒a^n-b^n= k(a-b)
where, k =[(a−b)n−1+C1n(a−b)n−2b+...+Cn−1n bn−1] [(a − b)^{n-1} + C_1^n (a - b)^{n-2} b+...+C_{n-1}^n\,b^{n-1}] is a natural number
This shows that (a - b) is a factor of (an−bn)(a^n - b^n), where n is a positive integer