Eliminate $x, y, z$ from the equations $(y + z)^{2} = 4 a^{2} y z, (z + x)^{2} = 4 b^{2} z x, (x + y)^{2} = 4 c^{2} x y$ .

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Solution

By multiplying all three equations
$64 a^{2} b^{2} c^{2} x^{2} y^{2} z^{2} = {(y + z)}^{2} {(z + x)}^{2} {(x + y)}^{2}$
i.e., $(y + z) (z + x) (x + y) = \pm 8 a b c x y z$
or $\frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} + \frac{x}{y} + \frac{y}{x} + 2 = \pm 8 a b c$
$\frac{y}{z} + 2 + \frac{z}{y} = 4 a^{2}$
$\frac{z}{x} + 2 + \frac{x}{z} = 4 b^{2}$
$\frac{x}{y} + 2 + \frac{y}{x} = 4 c^{2}$
$∴ \pm 8 a b c = 4 a^{2} - 2 + 4 b^{2} - 2 + 4 c^{2} - 2 + 2$
$a^{2} + b^{2} + c^{2} \pm 2 a b c = 1$

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