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

Given : $$a^2=by+cz$$

Adding $$'ax'$$ on both sides, we get :

=> $$a^2+ax=ax+by+cz$$

=> $$a(a+x)=ax+by+cz$$

=> $$\frac{1}{a+x}=\frac{a}{ax+by+cz}$$

=> $$\frac{x}{a+x}=\frac{ax}{ax+by+cz}$$ ----------(i)

Similarly, $$\frac{y}{b+y}=\frac{by}{ax+by+cz}$$ ----------(ii)

and $$\frac{z}{c+z}=\frac{cz}{ax+by+cz}$$ ----------(iii)

Adding equations (i),(ii) and (iii)

=> $$\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$$ $$= \frac{ax}{ax+by+cz}+\frac{by}{ax+by+cz}+\frac{cz}{ax+by+cz}$$

= $$\frac{ax+by+cz}{ax+by+cz}=1$$

=> Ans - (A)