If $y^2+z^2=3 y z, z^2+x^2=8 z x, x^2+y^2=4 x y$. then the value of $\frac{y^2}{x z}+\frac{x z}{y^2}$ is
If $y^2+z^2=3 y z, z^2+x^2=8 z x, x^2+y^2=4 x y$. then the value of $\frac{y^2}{x z}+\frac{x z}{y^2}$ is
$2$
$3$
$4$
$5$
Solution
(c)
$
\begin{aligned}
& y^2+z^2=3 y z \Rightarrow \frac{y}{z}+\frac{z}{y}=3 \\
& z^2+x^2=8 z x \Rightarrow \frac{z}{x}+\frac{x}{z}=8 \\
& x^2+y^2=4 x y \Rightarrow \frac{x}{y}+\frac{y}{x}=4
\end{aligned}
$
From Eqs. (i) and (iii),
$
\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{x}{z}+\frac{z}{x}+\frac{y^2}{x z}+\frac{x z}{y^2}=12
$
$\therefore \frac{y^2}{x z}+\frac{x z}{y^2}=12-8=4\left[\right.$ from Eq. (ii), $\left.\frac{x}{z}+\frac{z}{x}=8\right]$