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If $x = \log_{a} bc, y = \log_{b} ca, z = \log_{c} ab$ then xyz =

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$x + y + z + 1$

$x + y + z + 3$

$x + y + z + 2$

$x + y + z$

answer is C.

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Detailed Solution

$\begin{array}{l} 1 + x = \log_{a}^{a} + \log_{a}^{bc} = \log_{a}^{abc} \\ \Rightarrow \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = \log_{abc}^{abc} = 1 \end{array}$

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