We have $9^{-1} \cdot 27^{-1 / 3} = \frac{1}{9} \cdot \frac{1}{\sqrt[3]{27}} = \frac{1}{9} \cdot \frac{1}{3} = \frac{1}{27}$.
So, the expression inside the radical becomes $729 \cdot \sqrt[3]{\frac{1}{27}} = 729 \cdot \frac{1}{3} = 243$.
Then, take the fourth
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