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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

$$\mathrm{0} \\ $$

Commented by Algoritm last updated on 11/Jun/20

solve

$$\mathrm{solve} \\ $$

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

a+b+c=0 ⇒a^2 =−ba−ca ⇒b^2 =−ab−bc c^2 =−ca−bc (1/(a^2 +a^2 +bc))+(1/(b^2 +b^2 +ca))+(1/(c^2 +c^2 +ab)) (l/(a^2 −ab−ca+bc)) +(1/(b^2 −ab−bc+ca)) +(1/(c^2 −ca−bc+ab)) (1/((a−b)(a−c)))+(1/((b−a)(b−c))) +(1/((c−a)(c−b))) ((b−c+c−a+a−b)/(−(a−b)(b−c)(c−a))) =(0/(−(a−b)(b−c)(c−a)))=0 ans

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

Commented by Algoritm last updated on 11/Jun/20

thanks

$$\mathrm{thanks} \\ $$

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