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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

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    $\begingroup$ use that $$x^2+2xy+y^2=xy$$ thus we have $$(x+y)^2=xy$$ and finaly $$x+y=\sqrt{xy}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 27 '15 at 14:31
  • $\begingroup$ Although, as others have shown, the intended answer is surely -2, I just wanted to point out that the only REAL numbers satisfying $x^2 + xy + y^2 = 0$ are $x=y=0$. $\endgroup$ – jbuddenh Jul 27 '15 at 15:33
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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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  • $\begingroup$ which formula/theorem were used in this working? $\endgroup$ – wuiyang Jul 27 '15 at 15:13
  • $\begingroup$ @wuiyang, I hope not many apart from proofwiki.org/wiki/Exponent_Combination_Laws $\endgroup$ – lab bhattacharjee Jul 27 '15 at 15:17
  • $\begingroup$ how do you let $r^2 + r + 1 = 0$ become $r^3 -1$, where does the $(r-1)$ come from? $\endgroup$ – wuiyang Jul 27 '15 at 15:24
  • $\begingroup$ @wuiyang, As $r^2+r+1=0$ and $r^3-1=(r-1)(r^2+r+1)$ for finite $r, r^3-1=0$ $\endgroup$ – lab bhattacharjee Jul 27 '15 at 15:25
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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 $$-1^{1338}-1^{669}=-2$$

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  • $\begingroup$ How do you move from $x^2+xy+y^2=0$ to $\frac xy=e^{\pm\frac{2i\pi}{3}}$? $\endgroup$ – K. Rmth Jul 27 '15 at 15:13
  • $\begingroup$ Solving a quadratic in $\frac xy$ and obtaining the two complex cube roots of $1$ $\endgroup$ – David Quinn Jul 27 '15 at 15:16
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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)

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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,

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

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