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?

First of all substitute $x$ with $py$
Observe $y^3-1=(y-1)(?)=0$
So, $y=w,w$ is a complex cube root of unity
Now $(x+p)^n-x^n-p^n=p^n((1+y)^n-1-y^n)$
Now $f(n)=(1+w)^n-1-w^n=(-w^2)^n-1-w^n$
which is $=-1-w^n-w^{2n}$ if $n$ is odd
If $3\mid n,f(n)=-1-1^n-1^{2n}=?$
Else
$w^n\ne1,f(n)=-\dfrac{1-(w^3)^n}{1-w^n}=0$
Check for $n=3m+1,3m+2$
