<s>[INST] 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
 [/INST] Since $$x^3+2x^2y+2xy^2+y^3=(x+y)(x^2-xy+y^2)+2xy(x+y)=$$
$$=(x+y)(x^2+xy+y^2)=(x+y)\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right),$$
we see that $x^3+2x^2y+2xy^2+y^3=0$ for $x+y=0$ only.
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