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Question Number 192160 by universe last updated on 09/May/23

$$\mathrm{if}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{three}\mathrm{distinct}\mathrm{complex}\mathrm{numbers}$$ $$\mathrm{such}\mathrm{that}\frac{\mathrm{x}}{\mathrm{y}-\mathrm{x}}+\frac{\mathrm{y}}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}-\mathrm{y}}=0\mathrm{then}$$ $$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\Sigma }\frac{{\mathrm{x}}^2}{{(\mathrm{y}-\mathrm{z})}^2}$$

Commented bymehdee42 last updated on 09/May/23

$$\mathrm{\Sigma }?$$

Commented byFrix last updated on 09/May/23

$$\frac{\mathrm{x}}{\mathrm{y}-\mathrm{x}}+\frac{\mathrm{y}}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}-\mathrm{y}},\mathrm{no}\mathrm{symmetry},\mathrm{really}?$$

Commented byFrix last updated on 09/May/23

$$\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0$$ $$\Rightarrow$$ $$\frac{x^2}{{(y-z)}^2}+\frac{y^2}{{(z-x)}^2}+\frac{z^2}{{(x-y)}^2}=2$$

Answered by York12 last updated on 09/May/23

$$\frac{x-z}{y-x}=\frac{y}{x-z}\to {(x-z)}^2=y(y-x)$$ $$\frac{x-z}{y-x}+\frac{\underset{-}{+}\sqrt{y}}{\underset{-}{+}\sqrt{y-x}}=\frac{\underset{-}{+}2\sqrt{y}}{\underset{-}{+}\sqrt{y-x}}=0\to y=0$$ $$\therefore x=z$$ $$\therefore \sum _{sym}\frac{x^2}{{(y-z)}^2}=2$$

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