2) a) \((-\frac{16x^2}{27y^3})^2 \cdot (\frac{9y^2}{8x^2})^3\) ;

2) a) \((-\frac{16x^2}{27y^3})^2 \cdot (\frac{9y^2}{8x^2})^3 = \frac{256x^4}{729y^6} \cdot \frac{729y^6}{512x^6} = \frac{256 \cdot 729 \cdot x^4 \cdot y^6}{729 \cdot 512 \cdot y^6 \cdot x^6} = \frac{256}{512} \cdot \frac{x^4}{x^6} = \frac{1}{2x^2}\)

Ответ: \(\frac{1}{2x^2}\)