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.

We can use the binomial theorem for the coefficients of $(a+b)^7$.
The coefficients are 1,7,21,35,35,21,7,1, but we minus the $a^7$ and $b^7$.
So we have $7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6$, and then, let $a=0$, and since the result is 0 when we insert it, we have $a$ as a root. $b$ and $a+b$ works too. We now have $7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)$, which equals $7ab(a+b)(a^2+ab+b^2)^2$.
