2023/029) Given $x^2+xy+ y^2=0$ find the value of $(\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}$

We have $x^2+xy+ y^2= 0\cdots(1)$

 if x is zero the y is zero then x+y = 0 which is not possible as x+y is in denominator of the resultnat expression

Let $y=x\omega$

putting in (1) we get

$\omega ^2 + \omega + 1=0\cdots(2)$

Hence $\omega^3=1\cdots(2)$

Now $x+y = x + x\omega = x(1+\omega) = x(-\omega^2)= - x\omega^2$ using(1)

hence $\frac{x}{x+y} = \frac{x}{- x\omega^2} = - \omega\cdots(3)$

also  $\frac{y}{x+y} = \frac{x\omega}{- x\omega^2} = - \frac{1}{\omega} = \omega^2\cdots(4)$

hence  $(\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}) = \omega^{2023} + (\omega^2)^{2023}$

= $- \omega^{2023}  - \omega^{4036}$

=  $- \omega^{3  * 677 + 1} - \omega^{3 * 1354 + 2 }$

=  $- \omega  - \omega^2 $ using (3) 

= 1 using (2)