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.
 A: 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
$$\begin{align}
&a^7 + b^7 + c^7 = (C^2-B^3)A + 3B^2C - 2CB(A^2 - 2B) = 7B^2C\\
\implies &
(a+b)^7 - a^7 - b^7 = -7B^2C = 7ab(a+b)(a^2+ab+b^2)^2
\end{align}
$$
A: expanding the left-hand side we get
$$7\,{a}^{6}b+21\,{a}^{5}{b}^{2}+35\,{a}^{4}{b}^{3}+35\,{a}^{3}{b}^{4}+
21\,{a}^{2}{b}^{5}+7\,a{b}^{6}
$$
and the right-hand side:
$$
7\,{a}^{6}b+21\,{a}^{5}{b}^{2}+35\,{a}^{4}{b}^{3}+35\,{a}^{3}{b}^{4}+
21\,{a}^{2}{b}^{5}+7\,a{b}^{6}
$$
A: As  Mr. Chip suggested there’s no need to drag around $a$ and $b$ here. 
So I'm gonna proceed in this fashion:
$$(1+x)^7-1^7-x^7=7x(1+x)(1+x+x^2)^2$$
Now we divide by $(1+x)$ or we can factor it out:
Note that various divisions like that may call into question the rigor here. But it should be as easy as pie for you (I'm leaving it for you).
Well, we go on
$$(1+x)^6-(1-x+x^2-x^3+x^4-x^5+x^6)=7x(1+x+x^2)^2$$
$$7x+14x^2+21x^3+14x^4+7x^5=7x(1+x^2+x^4+2x+2x^2+2x^3)$$
$$7x(1+2x+3x^2+2x^3+x^4)=7x(1+2x+3x^2+2x^3+x^4)$$
So we got our result slightly faster this way. 
My point is you should be very proficient even when dealing with complicated expressions including the knoweledge of Multinomial theorem, which I used here on autopilot for getting the square of the trinomial. And the formula for $1\pm x^n$ and all the related subtleties are supposed to be easy when we talk about Olympiads. Seeing here $1^7+x^7$ should ring a bell that we can factor out $1+x\,.\,$ Hope my advice might be of some use.
A: Let $I = (a+b)^7-a^7-b^7$.
$$a^7+b^7 = (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$$
and
$$(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$
Then
\begin{eqnarray*}
  I &=& (a+b)(7a^5b+14a^4b^2+21a^3b^3+14a^2b^4+7ab^5)\\
  &=&7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\
  & =& 7ab(a+b)(a^2+ab+b^2)^2
\end{eqnarray*}
A: 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 $7ab(a+b)(a^2+ab+b^2)^2$.
