Ta có: \(Q=3 \log _{a} b+6 \cdot \frac{1}{2} \log _{a} b=6 \log _{a} b\).

\(\begin{aligned} P & =\frac{\log _{a} a^{3}+\log _{a} b^{2}-\left(\log _{b} b^{3}-\log _{b} a^{2}\right)}{\log _{a}^{2} b+1} \\ \text { Ta có: } \quad & \frac{3+2 \log _{a} b-3+2 \log _{b} a}{\log _{a}^{2} b+1}=\frac{2\left(\log _{a} b+\frac{1}{\log _{a} b}\right)}{\log _{a}^{2} b+1} \\ & =\frac{2\left(\frac{\log _{a}^{2} b+1}{\log _{a} b}\right)}{\log _{a}^{2} b+1}=\frac{2}{\log _{a} b}=2 \log _{b} a .\end{aligned}\)