Question from Spivak's Calculus. Hint makes no sense. I'm currently working through Spivak. This question has me a little bit tied up in knots. I was able to answer it, but I can't make heads or tails of the hint.
Here's the question:
Find out when $(x+y)^5=x^5+y^5$. Hint: from the assumption $(x+y)^5=x^5+y^5$ you should be able to derive the equation $x^3 + 2x^2y + 2xy^2+y^3= 0$ if $xy\neq 0$. (I got this part, the next part is what confuses me.) This implies that $(x+y)^3=x^2y + xy^2=xy(x+y)$. 
The last part of the above is what has me confused. What he's stating to be equal to $(x+y)^3$looks to me to be the difference of the binomial expansion of $(x+y)^3$ and the other equation given in the hint. Am I missing something???
 A: $$(x+y)^3 = x^3+3x^2y+3xy^2+y^3.$$
Suppose $x^3 + 2x^2y+2xy^2+y^3 = 0$. Add to both sides $x^2y+xy^2$:
$$x^3+2x^2y+2xy^2+y^3 + \color{red}{x^2y+xy^2} = \color{red}{x^2y+xy^2} \\
x^3+3x^2y+3xy^2+y^3 = x^2y+xy^2 \\
(x+y)^3 = xy(x+y).$$
A: As you pointed out, he is equating what he got from simplifying $(x+y)^5=x^5+y^5$ and the binomial expansion of $(x+y)^3$.
The first one is an equation, of which you're trying to find the solution. The second is an identity, which is always valid. The combination of both gives the said condition.
Explicitly, $(x+y)^3 = x^3+3x^2y+3xy^3+y^3 = x^3 + 2x^2y+2xy^2+y^3 + (x^2y+xy^2) = 0 + x^2y+xy^2 \\ = \,...$
A: The hint is not all that useful.
The hidden subtext of the problem is that for the first few odd $n$, the expressions $((x+y)^n - (x^n + y^n))$ factorize into easily understood pieces of degree $1$ and $2$.
Spivak is essentially asking for the factorization when $n=5$.  In addition to the obvious $x$ and $y$ there is a 

 factor of $(x+y)$ that can be verified by calculation or derived by substituting $y=-x$

and that uniquely determines the rest.
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