\(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)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 = (\(a^{-\frac{1}{6}})^3\) = \(a^{-\frac{1}{2}}\)