If a, b, c are three distinct positive number | vedclass

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(A) Given the expression: $[(\log_b a)(\log_c a) - 1] + [(\log_a b)(\log_c b) - 1] + [(\log_a c)(\log_b c) - 1] = 0$ Since $\log_a a = \log_b b = \log

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(A) Given the expression: $[(\log_b a)(\log_c a) - 1] + [(\log_a b)(\log_c b) - 1] + [(\log_a c)(\log_b c) - 1] = 0$ Since $\log_a a = \log_b b = \log_c c = 1$,the equation becomes: $(\log_b a)(\log_c a) + (\log_a b)(\log_c b) + (\log_a c)(\log_b c) = 3$ Let $x = \ln a, y = \ln b, z = \ln c$. The equation is: $(\frac{x}{y})(\frac{x}{z}) + (\frac{y}{x})(\frac{y}{z}) + (\frac{z}{x})(\frac{z}{y}) = 3$ $\frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} = 3$ Multiplying by $xyz$: $x^3 + y^3 + z^3 = 3xyz$ This identity holds if $x + y + z = 0$ or $x = y = z$. Since $a, b, c$ are distinct,$x, y, z$ are distinct,so $x + y + z = 0$. $\ln a + \ln b + \ln c = 0$ $\ln(abc) = 0$ $abc = e^0 = 1$.

If $a, b, c$ are three distinct positive numbers 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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