If y^2+z^2=a y z,z^2+x^2=b x z,and x^2+y^2=c | vedclass

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(C) Given equations are:$y^2+z^2=a y z \Rightarrow \frac{y}{z}+\frac{z}{y}=a$   $(i)$$z^2+x^2=b z x \Rightarrow \frac{z}{x}+\frac{x}{z}=b$   $(ii)$$x^

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$a c-b$

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(C) Given equations are:$y^2+z^2=a y z \Rightarrow \frac{y}{z}+\frac{z}{y}=a$   $(i)$$z^2+x^2=b z x \Rightarrow \frac{z}{x}+\frac{x}{z}=b$   $(ii)$$x^2+y^2=c x y \Rightarrow \frac{x}{y}+\frac{y}{x}=c$   $(iii)$Multiplying $(i)$ and $(iii)$:$(\frac{y}{z}+\frac{z}{y})(\frac{x}{y}+\frac{y}{x}) = ac$$\frac{x}{z} + \frac{y^2}{zx} + \frac{zx}{y^2} + \frac{z}{x} = ac$From $(ii)$,we know $\frac{z}{x}+\frac{x}{z}=b$. Substituting this into the expression:$b + \frac{y^2}{zx} + \frac{zx}{y^2} = ac$$\frac{y^2}{zx} + \frac{zx}{y^2} = ac-b$

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

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