If y^2+z^2=3yz,z^2+x^2=8zx,and x^2+y^2=4xy,th | vedclass

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(C) Given equations are:$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$  $(i)$$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$  $(ii)$$x^2+y^2=

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$4$

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(C) Given equations are:$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$  $(i)$$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$  $(ii)$$x^2+y^2=4xy \Rightarrow \frac{x}{y}+\frac{y}{x}=4$  $(iii)$Multiplying $(i)$ and $(iii)$:$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right) = \frac{xy}{zy} + \frac{y^2}{xz} + \frac{z}{y} \cdot \frac{x}{y} + \frac{z}{y} \cdot \frac{y}{x} = \frac{x}{z} + \frac{y^2}{xz} + \frac{xz}{y^2} + \frac{z}{x} = 12$$\left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$Substituting the value from $(ii)$:$8 + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$$\frac{y^2}{xz}+\frac{xz}{y^2} = 12 - 8 = 4$

If $y^2+z^2=3yz$,$z^2+x^2=8zx$,and $x^2+y^2=4xy$,then the value of $\frac{y^2}{xz}+\frac{xz}{y^2}$ is

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