find $\left( \frac{x}{x+y} \right)^{2007} + \left( \frac{y}{x+y} \right)^{2007}$

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 $(1)$

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 $$-1^{1338}-1^{669}=-2$$

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)

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,

$$\left( \frac{x}{y} \right)^3 = 1 $$