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?

Another approach:  there is a general factoring pattern for a sum of $n$th powers that works only when $n$ is odd.  Specifically, in the case $n=7$ we have
$$x^7 + y^7 = (x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)$$
Since both $(x+y)^7$ and $x^7 + y^7$ contain $(x+y)$ as a factor, we can write
$$(x+y)^7 - (x^7+y^7)= (x+y)\left((x+y)^6 - (x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)\right)$$
Now we can also expand $(x+y)^6$ using the binomial theorem:
$$(x+y)^6 = x^6 + 6x^5y + 15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$$
Plugging this into the equation directly above it, and combining like terms, we have
$$(x+y)^7 - (x^7+y^7)= (x+y)\left((x^6 + 6x^5y + 15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6) - (x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)\right) = (x+y)\left( 7x^5y + 14x^4y^2+21x^3y^3+14x^2y^4+7xy^5 \right)$$
Now in the second parenthesis, each term is a multiple of $7xy$, so factor that out too:
$$(x+y)^7 - (x^7+y^7)= (x+y)(7xy)\left(x^4 + 2x^3y+3x^2y^2+2xy^3+y^4 \right)$$
The last step is to factor $x^4 + 2x^3y+3x^2y^2+2xy^3+y^4 $.  This is maybe the trickiest part, as you have to somehow either recognize the factoring pattern
$$x^4 + 2x^3y+3x^2y^2+2xy^3+y^4 = (x^2 + xy + y^2)^2$$
which can be confirmed by expanding the right-hand side; if you didn't know this pattern already, I'm not sure how one could discover it other than by trial and error.
Anyway, now you can put it all together:
$$(x+y)^7 - (x^7+y^7)= (x+y)(7xy) (x^2 + xy + y^2)^2 $$
and you're done.
