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Question Number 98035 by Algoritm last updated on 11/Jun/20

Commented by som(math1967) last updated on 11/Jun/20

$$0$$

Commented by Algoritm last updated on 11/Jun/20

$$\mathrm{solve}$$

Commented by som(math1967) last updated on 11/Jun/20

$$\mathrm{a}+\mathrm{b}+\mathrm{c}=0$$$$\Rightarrow {\mathrm{a}}^2=-\mathrm{ba}-\mathrm{ca}$$$$\Rightarrow {\mathrm{b}}^2=-\mathrm{ab}-\mathrm{bc}{\mathrm{c}}^2=-\mathrm{ca}-\mathrm{bc}$$$$\frac{1}{{\mathrm{a}}^2+{\mathrm{a}}^2+\mathrm{bc}}+\frac{1}{{\mathrm{b}}^2+{\mathrm{b}}^2+\mathrm{ca}}+\frac{1}{{\mathrm{c}}^2+{\mathrm{c}}^2+\mathrm{ab}}$$$$\frac{\mathrm{l}}{{\mathrm{a}}^2-\mathrm{ab}-\mathrm{ca}+\mathrm{bc}}+\frac{1}{{\mathrm{b}}^2-\mathrm{ab}-\mathrm{bc}+\mathrm{ca}}$$$$+\frac{1}{{\mathrm{c}}^2-\mathrm{ca}-\mathrm{bc}+\mathrm{ab}}$$$$\frac{1}{(\mathrm{a}-\mathrm{b})(\mathrm{a}-\mathrm{c})}+\frac{1}{(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{c})}$$$$+\frac{1}{(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})}$$$$\frac{\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}$$$$=\frac{0}{-(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}=0\mathrm{ans}$$

Commented by Algoritm last updated on 11/Jun/20

$$\mathrm{thanks}$$

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