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If $y^2+z^2=a y z, z^2+x^2=b x z, x^2+y^2=c x y$, then the value of $\frac{x z}{y^2}+\frac{y^2}{z x}$ is
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The correct answer is:
$a c-b$
$\because y^2+z^2=a y z$
$\Rightarrow \quad \frac{y^2+z^2}{y z}=\frac{a y z}{y z}$
$\Rightarrow \quad \frac{y}{z}+\frac{z}{y}=a$ ...(i)
$\because \quad z^2+x^2=b z x$
$\Rightarrow \quad \frac{z}{x}+\frac{x}{z}=b$ ...(ii)
and $\quad x^2+y^2=c x y$
$\Rightarrow \quad \frac{x}{y}+\frac{y}{x}=c$ ...(iii)
Multiplying Eqs. (i) and (iii),
$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right)=a c$
$\Rightarrow \frac{x}{z}+\frac{y^2}{z x}+\frac{z x}{y^2}+\frac{z}{x}=a c$ ...(iv)
On subtracting Eq. (ii) from Eq. (iv),
$\left(\frac{x}{z}+\frac{z}{x}+\frac{y^2}{z x}+\frac{z x}{y^2}\right)-\left(\frac{z}{x}+\frac{x}{z}\right)=a c-b$
$\Rightarrow \frac{y^2}{z x}+\frac{z x}{y^2}=a c-b$
$\Rightarrow \quad \frac{y^2+z^2}{y z}=\frac{a y z}{y z}$
$\Rightarrow \quad \frac{y}{z}+\frac{z}{y}=a$ ...(i)
$\because \quad z^2+x^2=b z x$
$\Rightarrow \quad \frac{z}{x}+\frac{x}{z}=b$ ...(ii)
and $\quad x^2+y^2=c x y$
$\Rightarrow \quad \frac{x}{y}+\frac{y}{x}=c$ ...(iii)
Multiplying Eqs. (i) and (iii),
$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right)=a c$
$\Rightarrow \frac{x}{z}+\frac{y^2}{z x}+\frac{z x}{y^2}+\frac{z}{x}=a c$ ...(iv)
On subtracting Eq. (ii) from Eq. (iv),
$\left(\frac{x}{z}+\frac{z}{x}+\frac{y^2}{z x}+\frac{z x}{y^2}\right)-\left(\frac{z}{x}+\frac{x}{z}\right)=a c-b$
$\Rightarrow \frac{y^2}{z x}+\frac{z x}{y^2}=a c-b$
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