<s>[INST] find $\left( \frac{x}{x+y} \right)^{2007} + \left( \frac{y}{x+y} \right)^{2007}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

if $x, y$ are non-zero numbers satisfying $x^2 + xy + y^2 = 0$, find the value of $$\left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007}$$
(A). $2$ (B). $1$ (C). $0$ (D). $-1$ (E). $-2$

expanding it would give us $$ \frac { x^{2007} + y^{2007}} {(x+y)^{2007}}$$
how do I calculate this? Very appreciate for all of those who had helped me
 [/INST] Set $x=ry$
$\implies y^2(r^2+r+1)=0\implies r^2+r+1=0\implies r^3-1=(r-1)(r^2+r+1)=0$
$\implies r^3=1\ \  \  \ (1)$
$\dfrac x{x+y}=\dfrac{ry}{y+ry}=\dfrac r{1+r}$
$\dfrac y{x+y}=\dfrac y{y+ry}=\dfrac 1{1+r}$
As $2007\equiv3\pmod6=6a+3$ where $a=334$( in fact $a$ can be any integer)
The required sum $=\dfrac{r^{6a+3}+1}{(1+r)^{6a+3}}=\dfrac{(r^3)^{2a+1}+1}{(-r^2)^{6a+3}}=\dfrac{(r^3)^{2a+1}+1}{-(r^3)^{2(2a+1)}}$
Use $(1)$
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