# How do you evaluate 8^(-2/3)* 64^(1/6)?

Jul 9, 2015

Let's first look at the ${64}^{\frac{1}{6}}$

#### Explanation:

Since $64 = {8}^{2}$ we might as well write:

${8}^{- \frac{2}{3}} \cdot {\left({8}^{2}\right)}^{\frac{1}{6}} =$${8}^{- \frac{2}{3}} \cdot {8}^{2 \cdot \frac{1}{6}} =$

${8}^{- \frac{2}{3}} \cdot {8}^{\frac{1}{3}} =$${8}^{- \frac{2}{3} + \frac{1}{3}} = {8}^{- \frac{1}{3}}$

This may also be written as $\frac{1}{\sqrt[3]{8}}$