If log_a b + log_b c + log_c a vanishes,where | vedclass

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(A) Let $x = \log_a b$,$y = \log_b c$,and $z = \log_c a$.Given that $x + y + z = 0$.We know the algebraic identity: if $x + y + z = 0$,then $x^3 + y^3

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an odd prime

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(A) Let $x = \log_a b$,$y = \log_b c$,and $z = \log_c a$.Given that $x + y + z = 0$.We know the algebraic identity: if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Now,consider the product $xyz = (\log_a b) \cdot (\log_b c) \cdot (\log_c a)$.Using the change of base formula,$\log_a b = \frac{\ln b}{\ln a}$,$\log_b c = \frac{\ln c}{\ln b}$,and $\log_c a = \frac{\ln a}{\ln c}$.Thus,$xyz = \left( \frac{\ln b}{\ln a} \right) \cdot \left( \frac{\ln c}{\ln b} \right) \cdot \left( \frac{\ln a}{\ln c} \right) = 1$.Therefore,$x^3 + y^3 + z^3 = 3(1) = 3$.Since $3$ is an odd prime number,the correct option is $A$.

If $\log_a b + \log_b c + \log_c a$ vanishes,where $a, b,$ and $c$ are positive real numbers different from unity,then the value of $(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3$ is

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