If frac{m-a^{2}}{b^{2}+c^{2}}+frac{m-b^{2}}{c | vedclass

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(B) Given equation: $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}}{a^{2}+b^{2}}=3$Subtract $1$ from each term on the left sid

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$a^{2}+b^{2}+c^{2}$

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(B) Given equation: $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}}{a^{2}+b^{2}}=3$Subtract $1$ from each term on the left side:$(\frac{m-a^{2}}{b^{2}+c^{2}}-1)+(\frac{m-b^{2}}{c^{2}+a^{2}}-1)+(\frac{m-c^{2}}{a^{2}+b^{2}}-1)=3-3$$\frac{m-a^{2}-b^{2}-c^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}-c^{2}-a^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}-a^{2}-b^{2}}{a^{2}+b^{2}}=0$Let $K = m-(a^{2}+b^{2}+c^{2})$. Then the equation becomes:$K(\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{1}{a^{2}+b^{2}})=0$Since the sum of fractions is generally non-zero,we must have $K=0$.Therefore,$m-(a^{2}+b^{2}+c^{2})=0$,which implies $m=a^{2}+b^{2}+c^{2}$.

If $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}}{a^{2}+b^{2}}=3,$ then the value of $m$ is

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