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 would be to note that the function $ \ z \ =  \ (x+y)^7-(x^7+y^7) \ $ has an "exchange symmetry" between its variables $ \ x \ $ and $ \ y \ \ , $ which makes the function symmetric about the plane $ \ y \ = \ x \ \ . $  So its (bivariate) polynomial expression will have factors with this type of symmetry, such as $ \ x + y \ $ and $ \ xy \ \ . $
The Freshman's Dream notwithstanding, the binomial expansion applied to the expression will produce terms of the form $ \ C·x^m·y^n \ $ where $ \ m + n \ = \ 7 \ $ and with all of the binomial coefficients $ \  \frac{7!}{m!n!} \ \ , $ $ m \ \neq \ 0 \ , \ 7 \ \ , \  $  being divisible by $ \ 7 \ $ since that is prime.  So we can immediately see a factoring
$$ 7x^6y \ + \ 21x^5y^2 \ + \ 35x^4y^3 \ + \ 35x^3y^4 \ + \ 21x^2y^5 \ + \ 7xy^6 $$ $$ = \ \ 7 \ · \ xy \ · \ ( \ x^5 \ + \ 3x^4y \ + \ 5x^3y^2 \ + \ 5x^2y^3 \ + \ 3xy^4 \ + \ y^5 \ ) \ \ . $$
The "interchangeability" of the variables also permits us to see that $ \ ( x + y ) \ $ may be a factor:
$$  x^5 \ + \ 3x^4y \ + \ 5x^3y^2 \ + \ 5x^2y^3 \ + \ 3xy^4 \ + \ y^5 $$
$$ = \ \ x \ · \ ( \ x^4 \ + \ a·x^3y \ + \ b·x^2y^2 \ + \ axy^3 \ + \ y^4 \ ) $$ $$ \ + \ ( \ x^4 \ + \ a·x^3y \ + \ b·x^2y^2 \ + \ axy^3 \ + \ y^4 \ ) \ · \ y  \ \ . $$
Comparison of the coefficients requires that $ \ a + 1 \ = \ 3 \ $ and $ \ a + b \ = \ 5 \ \ , $ which can be solved consistently.  We then arrive at
$$ 7 \ · \ xy \ · \ (x + y) \ · \ ( \ x^4 \ + \ 2x^3y \ + \ 3x^2y^2 \ + \ 2xy^3 \  + \ y^4 \ ) \ \ . $$
The next most simple factor of suitable symmetry would be $ \ x^2 + y^2 \ \ ; $ no higher powers of $ \ (xy)^k \ $ can be "extracted".  Since there are terms in this remaining polynomial factor which have odd powers of $ \ x \ $ or $ \ y \ \ , $ however, this is a bit too simple.  We can instead next try factors of the form $   \ x^2 + cxy + y^2 \ \ . \ $  Although the symmetrical ("palindromic") arrangement of the coefficients strongly hints at what will happen, we will naïvely use two "distinct" factors to write
$$ x^4 \ + \ 2x^3y \ + \ 3x^2y^2 \ + \ 2xy^3 \  + \ y^4 \ \ =^{?} \ \ ( \ x^2 + cxy + y^2 \ ) · ( \ x^2 + dxy + y^2 \ ) $$
$$ =  \ \ ( \ x^4 + cx^3y + x^2y^2 \ ) \ + \  ( \ dx^3y + cdx^2y^2 + dxy^3 \ ) \ + \  ( \ x^2y^2 + cxy^3 + y^4 \ ) \ \ . $$
This factorization succeeds, as $ \ c + d \ = \ 2 \ $ and $ \ 1 + cd + 1 \ = \ 3 \ \ $ can also be solved consistently by $ \ c \ = \ d \ = \ 1 \ \ . $ So we find the two factors to be identical.
We thus obtain the factorization
$$ (x \ + \ y)^7 \ - \ (x^7 \ + \ y^7) \ \ = \ \ 7 \ · \ xy \ · \ (x + y) \ · \ ( \ x^2 + xy + y^2 \ )^2 \ \ . $$
