It is given that $n$ is an odd integer greater than $3$ but not a multiple of $3$. Prove that $x^3 +x^2 +x$ is a factor of $(x+1)^n -x^n -1$.
I tried to write $x^3+x^2+x=x(x^2+x+1) =x(x-\omega^2)(x-\omega)$ where $\omega$ and $\omega^2$ are cube roots of unity $≠1$. Then I tried to do it by putting values but could not get the answer
Modulo $x$, $(x+1)^n-x^n-1\equiv 1^n-0-1=0$, so $(x+1)^n-x^n-1$ is divisible by $x$.
It remains to be shown that $(x+1)^n-x^n-1$ is divisible by $x^2+x+1$. Well,
$(\omega+1)^n-\omega^n-1=(-\omega^2)^n-\omega^n-1=-(\omega^{2n}+\omega^n+1)$ for $n$ odd.
Furthermore, for $n$ not a multiple of $3$, $(\omega^{2n}+\omega^n+1)(\omega^{n}-1)=\omega^{3n}-1=0$ and $\omega^n-1\ne0$,
so $\omega^{2n}+\omega^n+1=0$. Therefore, $\omega$ is a zero of $(x+1)^n-x^n-1$.
