xxn−1−nαn−1+αn(n−1) is divisible by  (x−α)2 for

Let P(n)=xxn−1−nαn−1+αn(n−1)=(x−α)2g(x)P(1)≡00istrue.Let xxk−1−kαk−1+αk(k−1)=(x−α)2g(x)P(k+1)≡xxk−(k+1)αk+αk+1(k)≡(x−α)2xg(x)+kαk−1So, holds for all n∈N