Question : 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:
Option 1: $1$
Option 2: $a+b+c$
Option 3: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Option 4: $0$
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Correct Answer: $1$
Solution : Given: $a^{2}=by+cz$ $b^{2}=cz+ax$ $c^{2}=ax+by$ Then, value of $\frac{x}{a\:+\:x}+\frac{y}{b\:+\:y}+\frac{z}{c\:+\:z}$ $= \frac{a\:×\:x}{a\:×\:(a\:+\:x)}+\frac{b\:×\:y}{b\:×\:(b\:+\:y)}+\frac{c\:×\:z}{c\:×\:(c\:+\:z)}$ $= \frac{ax}{(a^{2}\:+\:ax)}+\frac{by}{(b^{2}\:+\:by)}+\frac{cz}{(c^{2}\:+\:cz)}$ Substituting the value of $a^{2}, b^{2}, c^{2}$ in this expression, ⇒ $\frac{ax}{(by\:+\:cz\:+\:ax)}+\frac{by}{(cz\:+\:ax\:+\:by)}+\frac{cz}{(ax\:+\:by\:+\:cz)}$ Taking LCM of denominators, we have: ⇒ $\frac{ax\:+\:by\:+\:cz}{(ax\:+\:by\:+\:cz)} = 1$ Hence, the correct answer is 1.
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