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

Explanation

We can write `a^n` as
`a^n=[a-b+b]^n`
`a^n=[b+(a-b)]^n`

We know that

`(a+b)^n=^nC_0a^n+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

Putting `a=b` and `b=a-b`

`therefore`
`a^n=[b+(a-b)]^n=^nC_0b^n+^nC_1b^(n-1)(a-b)+^nC_2b^(n-2)(a-b)^2+.......+^nC_n(a-b)^n`

`(a)^n=b^n+^nC_1b^(n-1)(a-b)+^nC_2b^(n-2)(a-b)^2+.......+(a-b)^n`

`a^n-b^n=^nC_1b^(n-1)(a-b)+^nC_2b^(n-2)(a-b)^2+.......+(a-b)^n`

Taking `(a-b)` common

`a^n-b^n=(a-b)[ " ^nC_1b^(n-1)+^nC_2b^(n-2)(a-b)+.......+(a-b)^(n-1)]`

`a^n-b^n=(a-b)[k]`
` " " " " " " " " `where `k=[ " ^nC_1b^(n-1)+^nC_2b^(n-2)(a-b)+.......+(a-b)^(n-1)]` is a natural number

Hence, `(a-b)` is a factor of `a^n-b^n`