Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$ 
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$

In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into very complicated expressions, and are most likely wrong. Hints or solutions are welcomed. Sorry if this is a duplicate.

Let $c = -(a+b)$. Consider following cubic polynomial having $a, b, c$ as roots:
$$t^3-At^2 + Bt - C \stackrel{def}{=} (t-a)(t-b)(t-c)
\quad\text{ where }\quad 
\begin{cases}
A &= a + b + c = 0\\
B &= ab+bc+ca = -(a^2+b^2+ab)\\
C &= abc = -ab(a+b)
\end{cases}$$
When $t$ is one of $a, b, c$, we have $t^3 = C - Bt$. This implies
$$t^7 = (C-Bt)^2 t = C^2 t - 2CB t^2 + B^2t^3
= (C^2 - B^3)t + B^2C -2CB t^2$$
Substitute $t$ by $a,b,c$ and sum over them. Together with
$$\begin{align}a + b + c 
&= A = 0\\
a^2 + b^2 +c^2 &= (a+b+c)^2 - 2(ab+bc+ca) = A^2 - 2B = -2B
\end{align}$$
we obtain
$$\begin{align}
&a^7 + b^7 + c^7 = (C^2-B^3)A + 3B^2C - 2CB(A^2 - 2B) = 7B^2C\\
\implies &
(a+b)^7 - a^7 - b^7 = -7B^2C = 7ab(a+b)(a^2+ab+b^2)^2
\end{align}
$$
