Binomial to the power of five, equality proof I want to find out when the equality $(x+y)^5=x^5+y^5$ for real numbers $y$ and $x$ holds.
Expanding this binomial yields $(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4$
Factoring the right side of the above equality gives $x^5+y^5+5xy(x^3+2x^2y+2xy^2+y^3)$
If $(x+y)^5=x^5+y^5$, then $5xy(x^3+2x^2y+2xy^2+y^3)=0$ which implies that
$xy=0$ or $x^3+2x^2+2xy^2+y^3=0$
if $xy\neq0$ then $x^3+2x^2y+2xy^2+y^3=0$ and thus $(x+y)^3=xy(x+y)$
This is where I'm stuck. How can the last equality help me to figure out when the original equality holds? I know that $xy\neq0$ and I can somehow see that showing that $x+y=0$ seems to be an appropriate step at this moment, but I'm now sure.
Thanks

It can be helpful with equations like  your last one there to gather terms on one side, and factor. Thus: $$(x+y)^3-xy(x+y)=0\\\implies (x+y)((x+y)^2-xy)=0\\ \implies x+y=0\,\,\,\text{ or }\,\,\, (x+y)^2-xy=0$$
What can you do with that second equation?
