Eliminate $x, y, z$ between the equations $y^{2} + z^{2} = a y z$ , $z^{2} + x^{2} = b z x$ , $x^{2} + y^{2} = c x y$ .

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Solution

$y^{2} + z^{2} = a y z$ ...... $(i)$
$z^{2} + x^{2} = b z x$ ...... $(i i)$
$x^{2} + y^{2} = c x y$ ...... $(i i i)$
Divide both sides by $y z$ in $(i)$ , we get
$\frac{y}{z} + \frac{z}{y} = a$ ...... $(i v)$
Similarly, $\frac{z}{x} + \frac{x}{z} = b$ ...... $(v)$ and
$\frac{x}{y} + \frac{y}{x} = c$ ...... $(v i)$
Multiply the three equations $(i v)$ , $(v)$ and $(v i)$ , we get
$2 + \frac{y^{2}}{z^{2}} + \frac{z^{2}}{y^{2}} + \frac{z^{2}}{x^{2}} + \frac{x^{2}}{z^{2}} + \frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} = a b c$
But ${(\frac{y}{z} + \frac{z}{y})}^{2} = a^{2}$ ...... From $(i v)$
$\frac{y^{2}}{z^{2}} + \frac{z^{2}}{y^{2}} + 2 = a^{2}$
Similarly, we can write
$\frac{z^{2}}{x^{2}} + \frac{x^{2}}{z^{2}} + 2 = b^{2}$
$\frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} + 2 = c^{2}$
$2 + a^{2} - 2 + b^{2} - 2 + c^{2} - 2 = a b c$
$a^{2} + b^{2} + c^{2} - 4 = a b c$
$a^{2} + b^{2} + c^{2} = a b c + 4$
Hence, it is eliminated.

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