Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.  

Factor $(x+y)^7-(x^7+y^7)$.

I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's 
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?

( Motivation for this (late) answer is to demistify the derivation of the square factor, and show that it reduces to the routine factorization of an associated quadratic. )
Using that $\,x^7 + 1 = (x+1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)\,$:
$$
(x+1)^7 - x^7 - 1 = (x+1)\left((x+1)^6 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1\right) \tag{1}
$$
The second factor is a palindromic polynomial, which suggests the substitution $\,u = x + \dfrac{1}{x}\,$. Then $\,x^2 + \dfrac{1}{x^2} = u^2 - 2\,$, $\,x^3 + \dfrac{1}{x^3} = u^3 - 3u\,$, and:
$$
\begin{align}
& (x+1)^6 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1
\\ =\; &(x^2 + 2x + 1)^3 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1
\\ =\; &x^3\left(\left(x+\frac{1}{x}+2\right)^3 - \left(x^3 + \frac{1}{x^3} \right)+\left(x^2 + \frac{1}{x^2}\right)-\left(x + \frac{1}{x}\right)+1\right)
\\ =\; &x^3 \left((u+2)^3 - (u^3 - 3u) + (u^2 - 2) - u + 1\right)
\\ =\; &x^3 \left(7 u^2 + 14 u + 7\right)
\\ =\; &7x^3(u+1)^2
\\ =\; &7x(ux+x)^2
\\ =\; &7x(x^2 + x + 1)^2 \tag{2}
\end{align}
$$
Piecing together $\,(1)\,$ and $\,(2)\,$:
$$
(x+1)^7 - x^7 - 1 = 7x (x+1) \left(x^2+x+1\right)^2
$$
Substituting $\,x \mapsto \dfrac{x}{y}\,$ and multiplying by $\,y^7\,$ gives the identity in OP's question.
