\(\frac{9a^2}{2a^3}\)
\(\frac{3}{2a^3}\)
\(\frac{2}{3a^2}\)
\(\frac{3a^2}{2}\)
Correct answer is B
\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\) = \(\frac{3}{2}\) x \(\frac{1}{a^3}\) = \(\frac{3}{2a^3}\)
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