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
 A: 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 $(1)$
A: Solving $$x^2+xy+y^2=0\Rightarrow \frac xy=e^{\pm\frac{2i\pi}{3}}\Rightarrow (\frac xy)^3=1$$
Since $x+y$ can be replaced with $-\frac{y^2}{x}$, the expression boils down to $$-1^{1338}-1^{669}=-2$$
A: Dividing $x^{2} + 2xy + y^{2}$ by $y^{2}$ gives an equation whose roots are the non-real cube roots of unity. That is, say $x/y$ = $\omega$ then $y/x$ = $\omega^{2}$. With $x + y = \sqrt{xy}$, the given equation can now be expressed conveniently in terms of these complex roots, and I think the individual terms will come out to 1 + 1 = 2 or -1 - 1 = -2 (I have not taken that trouble, sorry)
A: Hint: 
We have:
$$A = \left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007}$$
$$ = \left( \frac{x}{y} \right)^{1003} \frac{x}{x+y} + \left( \frac{y}{x} \right)^{1003}\frac{y}{x+y} $$
$$ = - \left[ \left( \frac{x}{y} \right)^{1002}+ \left( \frac{y}{x} \right)^{1002} \right]$$
From the condition $x^2 + xy + y^2 =0$, we have
 $$\left( \frac{x}{y} \right)^2 + \left( \frac{x}{y} \right) + 1 =0 $$
or,
$$\left( \frac{x}{y} \right)^3 = 1 $$
