Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).  

if $\omega$ is a complex cube root of unity, then both $\omega$ and $1+\omega$ are sixth roots of unity. hence for $k \ge 1$$\omega$ is a root of:
$$
(x+1)^{6k+1} - x^{6k+1} - 1 = 0 \tag{1}
$$
likewise for $\omega^2$
but these are exactly the roots of
$$
x^2+x+1=0
$$
this shows that $a^2+ab+b^2$ is a factor of $(a+b)^{6k+1}-a^{6k+1}-b^{6k+1}$ 
however the derivative of the LHS of (1) is $(6k+1)\left((x+1)^{6k}-x^{6k}\right)$
which is also satisfied by $\omega, \omega^2$, so each is, in fact, a double root of the equation (1). hence the result
