\[\frac{6 - 3a}{8a + 4b} \cdot \frac{4a^2 + 4ab + b^2}{a - 2} = \frac{-3(a - 2)}{4(2a + b)} \cdot \frac{(2a + b)^2}{a - 2} = -\frac{3}{4} \cdot (2a + b)\]
\[-\frac{3}{4} (2(6) + (-4)) = -\frac{3}{4} (12 - 4) = -\frac{3}{4} \cdot 8 = -6\]
Ответ: -6