If a, b, c are distinct positive numbers,each | vedclass

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(A) Let $x = \ln a, y = \ln b, z = \ln c$. Since $a, b, c 
eq 1$,$x, y, z 
eq 0$.Given equation: $[(\frac{x}{y})(\frac{x}{z}) - 1] + [(\frac{y}{x})(

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(A) Let $x = \ln a, y = \ln b, z = \ln c$. Since $a, b, c 
eq 1$,$x, y, z 
eq 0$.Given equation: $[(\frac{x}{y})(\frac{x}{z}) - 1] + [(\frac{y}{x})(\frac{y}{z}) - 1] + [(\frac{z}{x})(\frac{z}{y}) - 1] = 0$$\Rightarrow \frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} - 3 = 0$$\Rightarrow \frac{x^3 + y^3 + z^3}{xyz} = 3$$\Rightarrow x^3 + y^3 + z^3 - 3xyz = 0$Using the identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0$,we have either $x + y + z = 0$ or $x^2 + y^2 + z^2 - xy - yz - zx = 0$.If $x^2 + y^2 + z^2 - xy - yz - zx = 0$,then $\frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$,which implies $x = y = z$,but $a, b, c$ are distinct.Therefore,$x + y + z = 0$,which means $\ln a + \ln b + \ln c = 0$.$\Rightarrow \ln(abc) = 0 = \ln 1$.Thus,$abc = 1$.

If $a, b, c$ are distinct positive numbers,each different from $1$,such that $[(\log _b a)(\log _c a) - (\log _a a)] + [(\log _a b)(\log _c b) - (\log _b b)] + [(\log _a c)(\log _b c) - (\log _c c)] = 0$,then $abc =$

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