Cho $a>0, b>0$ thỏa mãn $\log _{16}(3 a+2 b)=\log _9 a=\log _{12}...

Đặt $t=\log _{16}(3 a+2 b)=\log _9 a=\log _{12} b$
$
\begin{aligned}
& \Rightarrow\left\{\begin{array}{l}
3 a+2 b=16^t \\
a=9^t \\
b=12^t
\end{array} \Rightarrow 3.9^t+2.12^t=16^t \Leftrightarrow 3\left(\dfrac{9}{16}\right)^t+2 .\left(\dfrac{3}{4}\right)^t=1\right. \\
& \Leftrightarrow\left[\begin{array}{l}
\left(\dfrac{3}{4}\right)^t=\dfrac{1}{3} \\
\left(\dfrac{3}{4}\right)^t=-1(v n)
\end{array} \Rightarrow \dfrac{a}{b}=\dfrac{1}{3} .\right.
\end{aligned}
$
Vậy $\dfrac{a^3-a b^2-b^3}{a^3+a^2 b+3 b^3}=\dfrac{\left(\dfrac{a}{b}\right)^3-\left(\dfrac{a}{b}\right)-1}{\left(\dfrac{a}{b}\right)^3+\left(\dfrac{a}{b}\right)^2+3}=-\dfrac{7}{17}$