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Question Number 169610 by mathlove last updated on 04/May/22

Answered by som(math1967) last updated on 04/May/22

$$\frac{{\alpha }^2}{{\alpha }^2+\alpha \beta +\alpha \gamma }+\frac{{\beta }^2}{{\beta }^2+\alpha \beta +\beta \gamma }+\frac{{\gamma }^2}{{\gamma }^2+\beta \gamma +\gamma \alpha }★$$$$=\frac{{\alpha }^2}{\alpha (\alpha +\beta +\gamma )}+\frac{{\beta }^2}{\beta (\alpha +\beta +\gamma )}+\frac{{\gamma }^2}{\gamma (\alpha +\beta +\gamma )}$$$$=\frac{(\alpha +\beta +\gamma )}{(\alpha +\beta +\gamma )}=1[\alpha +\beta +\gamma \ne 0]$$$$★\alpha \beta +\beta \gamma +\gamma \alpha =0$$$$\therefore -\beta \gamma =\alpha \beta +\alpha \gamma$$$$-\gamma \alpha =\alpha \beta +\beta \gamma$$$$-\alpha \beta =\beta \gamma +\gamma \alpha$$

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