1 over 9 to the power of 3 over 2

\((\frac19)^{\frac32}=\sqrt{\frac19^3}=\sqrt{\frac19\cdot\frac19\cdot\frac19}=\frac19\sqrt{\frac19}=\frac19\cdot\frac{\sqrt1}{\sqrt9}=\frac{1}{9}\cdot\frac13\mathbf{=\frac{1}{27}}\)

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