If a^{2}=by+cz, b^{2}=cz+ax, c^{2}=ax+by, the | vedclass

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(A) Given equations are:$a^{2}=by+cz$ ... $(1)$$b^{2}=cz+ax$ ... $(2)$$c^{2}=ax+by$ ... $(3)$Adding $ax$ to both sides of $(1)$:$a^{2}+ax = ax+by+cz$$

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$1$

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(A) Given equations are:$a^{2}=by+cz$ ... $(1)$$b^{2}=cz+ax$ ... $(2)$$c^{2}=ax+by$ ... $(3)$Adding $ax$ to both sides of $(1)$:$a^{2}+ax = ax+by+cz$$a(a+x) = ax+by+cz$$\frac{a}{a+x} = \frac{a^{2}}{ax+by+cz}$ (This is not the direct path,let's use the identity $\frac{x}{a+x} = 1 - \frac{a}{a+x}$)Actually,from $a(a+x) = ax+by+cz$,we have $\frac{a}{a+x} = \frac{a^{2}}{ax+by+cz}$.Alternatively,$\frac{x}{a+x} = \frac{ax}{a(a+x)} = \frac{ax}{ax+by+cz}$.Similarly,$\frac{y}{b+y} = \frac{by}{b(b+y)} = \frac{by}{ax+by+cz}$.$\frac{z}{c+z} = \frac{cz}{c(c+z)} = \frac{cz}{ax+by+cz}$.Adding these three expressions:$\frac{x}{a+x} + \frac{y}{b+y} + \frac{z}{c+z} = \frac{ax+by+cz}{ax+by+cz} = 1$.

If $a^{2}=by+cz, b^{2}=cz+ax, c^{2}=ax+by,$ then the value of $\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$ is

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