If a+b+c=0, then find the value of frac{1}{b^ | vedclass

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(A) Given $a+b+c=0.$ We need to find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$ Since $a+b+c=

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$0$

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(A) Given $a+b+c=0.$ We need to find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$ Since $a+b+c=0,$ we have $a+b=-c.$ Squaring both sides,$(a+b)^{2}=(-c)^{2},$ which gives $a^{2}+b^{2}+2ab=c^{2}.$ Rearranging,$b^{2}+c^{2}-a^{2} = b^{2}+(a^{2}+2ab+b^{2})-a^{2} = 2b^{2}+2ab = 2b(a+b).$ Since $a+b=-c,$ we get $b^{2}+c^{2}-a^{2} = 2b(-c) = -2bc.$ Similarly,$c^{2}+a^{2}-b^{2} = -2ac$ and $a^{2}+b^{2}-c^{2} = -2ab.$ Substituting these into the expression: $\frac{1}{-2bc} + \frac{1}{-2ac} + \frac{1}{-2ab} = \frac{-a-b-c}{2abc}.$ Since $a+b+c=0,$ the numerator is $0.$ Thus,the value of the expression is $0.$

If $a+b+c=0,$ then find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$

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