Cho \(\cot \alpha  = 3\).

* Hướng dẫn giải

\(\frac{{3\sin \alpha  - 2\cos \alpha }}{{12{{\sin }^3}\alpha  + 4{{\cos }^3}\alpha }} = \frac{{3\left( {\frac{1}{{{{\sin }^2}\alpha }}} \right) - 2\cot \alpha \left( {\frac{1}{{{{\sin }^2}\alpha }}} \right)}}{{12 + 4{{\cot }^3}\alpha }} = \frac{{\left( {3 - 2\cot \alpha } \right)\left( {1 + {{\cot }^2}\alpha } \right)}}{{12 + 4{{\cot }^3}\alpha }} =  - \frac{1}{4}\)