If `n` is n odd integer that is greater than or equal to 3 but not la ultiple of 3, then prove that `(x+1)^n=x^n-1` is divisible by `x^3+x^2+xdot`

Explanation

Let `f(x) = (x+1)^(n) - x^(n) -1`
`x^(2) + x^(2) + x = x(x^(2)+ x + 1) = x (x-omega)(x-omega^(2))`
`f(0) = (0 + 1)^(n)-0^(n) - 1 =0`
`f(omega) = (omega + 1)^(n) -omega^(n) -1 =(-omega^(2))^(n)-omega^(n) -1 = - (omega^(2n) + omega^(n) + 1) = 0`
When n is not mulitple of 3.
`f(omega^(2)) = (omega^(2) +1)^(2n) -1`
`= (-omega)^(n) - omega^(2n) - 1 = - (omega^(n) + omega^(2n) + 1) =0`
when n is not multipe of 3.