Math Book Cambodia: Prove that `x^2023+y^2023+z^2023=0` If `(x^2+y^2+z^2)/(a^2+b^2+c^2)=x^2/a^2+y^2/b^2+z^2/c^2`

Wednesday, March 22, 2023

Prove that $x^{2023} + y^{2023} + z^{2023} = 0$ $x^2023+y^2023+z^2023=0$ If $\frac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}$ $(x^2+y^2+z^2)/(a^2+b^2+c^2)=x^2/a^2+y^2/b^2+z^2/c^2$

This is the problem that I picked from Vietnamese collection problems which is the basic for kind of this question.
From that origin problem is $x^{2019} + y^{2019} + z^{2019} = 0$ $x^2019+y^2019+z^2019=0$ but, I have changed it to $2023$ $2023$ .

Problem in Khmer language:

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Another problem, you all can learn more.

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