<s>[INST] 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?
 [/INST] Users have approached the problem using the Binomial Theorem, namely,

$$(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k,\tag1$$

such that the formula for a binomial coefficient is expressed thus: $$\dbinom nk\stackrel{\small\text{def}}{=}\frac{n!}{k!(n-k)!}.\tag{$n!=1\times 2\times 3\times\cdots\times n$}$$ It follows, then, that by substituting $(a,b,n)=(x,y,7)$ respectively, the answer is as follows: $$\begin{align}(x+y)^7&=x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7 \\ &= x^7+y^7+7(x^6y+xy^6)+21(x^5y^2+x^2y^5)+35(x^4y^3+x^3y^4) \\ &=x^7+y^7+7\big(x^6y+xy^6+3(x^5y^2+x^2y^5)+5(x^4y^3+x^3y^4)\big) \\ &=x^7+y^7+7xy\big(x^5+y^5+3xy(x^4+y^4)+5xy(x^3+y^3)\big),\end{align}$$ to which we can then factor $-$ but there is a better way!

Of course now, $(x+y)^7-(x^7+y^7)$ is a multiple of $x+y$. Let $x^7+y^7=x^7-(-y)^7$, then you can use the following formula from my answer to this post in order to factor the equation:

$$a^n-b^n=(a-b)\sum_{k=1}^na^{n-k}b^{k-1}.\tag2$$

It follows, then, that $$(x+y)^7-(x^7+y^7)=(x+y)\bigg((x+y)^6-\sum_{k=1}^7x^{n-k}y^{k-1}\bigg).$$ The following step requires you to use the binomial theorem for the algebraic expression $(x+y)^6$, which is much easier than calculating for $(x+y)^7$ by hand.
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