If \(b^3\) = \(a^{-2}\) and \(c^{\frac{1}{3}}\) = \(a^\frac{1}{2}\)b, express c in terms of a
\(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b
b = \(a^{-2}\) and \(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b
b = \(a^{-\frac{2}{3}}\)
\(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\) x \(a^{-\frac{2}{3}}\) = \(a^{-\frac{1}{6}}\)
take the cube of both sides
c = (\(a^{-\frac{1}{6}})^3\) = \(a^{-\frac{1}{2}}\)
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