A problem with logarithms.

$\log_{b} \left ( \dfrac{a}{bc} \right ) = 1$

$\implies \dfrac{a}{bc} = b$

$\implies a = b^2c$

$\log_{b} \left(\dfrac{c}{ab} \right) = \log_{b} \left(\dfrac{c}{b^3 c} \right) = \log_{b} \left(\dfrac{1}{b^3} \right) = \boxed{-3}$

$\begin{aligned} \log_b \frac a{bc} & = 1 \\ \log_b a - \log_b b - \log_b c & = 1 \\ \log_b a - 1 - \log_b c & = 1 \\ \color{#3D99F6} \log_b a - \log_b c & \color{#3D99F6} = 2 & \color{#3D99F6} ...(1) \end{aligned}$

Then,

$\begin{aligned} \log_b \frac c{ba} & = \log_b c - \log_b b - \log_b a \\ & = \log_b c -1 - \log_b a \\ & = - \left({\color{#3D99F6}\log_b a - \log_b c}\right) - 1 & \small \color{#3D99F6} \text{Note }(1): \log_b a - \log_b c = 2 \\ & = - \left({\color{#3D99F6}2}\right) - 1 \\ & = \boxed{-3} \end{aligned}$

$log_{b} \left ( \frac{a}{bc} \right ) = log_{b} \left ( \frac{a}{c} \right ) + log_{b} \left ( \frac{1}{b} \right ) = log_{b} \left ( \frac{a}{c} \right ) - 1 = 1.$ $\therefore log_{b} \left ( \frac{a}{c} \right ) = 2.$ $log_{b} \left ( \frac{c}{a} \right ) = -2.$ $log_{b} \left ( \frac{c}{ab} \right ) = -2 - 1 = -3.$