\(\frac{r}{2}\)
2r
\(\frac{1}{2r}\)
\(\frac{2}{r}\)
Correct answer is B
\(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\) = \(((64r^{-6})^{1/_2})^{1/_3}\) =\((64r^{-6})^{1/_6}\)
=\(64)^{1/_6}(r^{-6})^{1/_6}\)
=2/r
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