Prove that $x^2+px+p^2$ is a factor $(x+p)^n-x^n-p^n$, if $n$ be odd and not divisible by $3$. Question:
Prove that $x^2+px+p^2$ is a factor of $(x+p)^n-x^n-p^n$, if $n$ is odd and is not divisible by $3$.
My approach:
$$(x+p)^n-x^n-p^n=\sum_{r=0}^n\limits {n\choose r} x^{n-r}p^r-x^n-p^n$$ What can I do after that?

You need to show every root of $x^2 + px + p^2$ is a root of $(x + p)^n - x^n - p^n$. The quadratic formula gives that $x = p\omega$ or $p\bar{\omega}$
where $w = -{1 \over 2} + {\sqrt{3} \over 2}i$ is a complex third root of unity. So what you need to show is that for the values of $n$ in question
that 
$$(p\omega + p)^n - (p\omega)^n - p^n = 0$$
You also need this for $\bar{\omega}$ in place of $\omega$ but that will follow by taking complex conjugates of this equation.
Hence you need to show for your values of $n$ that
$$(\omega + 1)^n = \omega^n + 1$$
I won't finish this off but you can reduce this to checking a small list of $n$.
