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

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(A) Given equation: $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b}=4(a+b+c)$Let $x = (a+b+c)^{2}$.Substituting $x$ in the equation:$\fra

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

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(A) Given equation: $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b}=4(a+b+c)$Let $x = (a+b+c)^{2}$.Substituting $x$ in the equation:$\frac{(a+b+c)^{2}-a^{2}}{b+c} + \frac{(a+b+c)^{2}-b^{2}}{c+a} + \frac{(a+b+c)^{2}-c^{2}}{a+b}$Using the identity $A^{2}-B^{2} = (A-B)(A+B)$:$= \frac{(a+b+c-a)(a+b+c+a)}{b+c} + \frac{(a+b+c-b)(a+b+c+b)}{c+a} + \frac{(a+b+c-c)(a+b+c+c)}{a+b}$$= \frac{(b+c)(2a+b+c)}{b+c} + \frac{(a+c)(a+2b+c)}{c+a} + \frac{(a+b)(a+b+2c)}{a+b}$$= (2a+b+c) + (a+2b+c) + (a+b+2c)$$= 4a+4b+4c = 4(a+b+c)$Since the expression satisfies the equation,$x = (a+b+c)^{2}$.

If $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b}=4(a+b+c),$ then $x$ is equal to:

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