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Saturday, June 24, 2023

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)



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