Question

how to solve cube root of 8^-2

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{cube}\mathrm{root}\mathrm{of}{8}^{-2}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \sqrt[3]{8^{-2}} \\ \mathrm{Use}\mathrm{the}\mathrm{formula}\sqrt[\mathrm{n}]{\mathrm{x}}={\mathrm{x}}^{1/\mathrm{n}}\mathrm{to}\mathrm{get}, \\ \sqrt[3]{8^{-2}}={\left(8^{-2}\right)}^{\frac{1}{3}} \\ \mathrm{Rewrite}8\mathrm{as}8=2\times 2\times 2=2^3\mathrm{to}\mathrm{get} \\ \sqrt[3]{8^{-2}}={\left({\left(2^3\right)}^{-2}\right)}^{\frac{1}{3}} \\ \mathrm{Use}\mathrm{the}\mathrm{rule}\mathrm{of}\mathrm{exponent}{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}\times \mathrm{n}} \\ \sqrt[3]{8^{-2}}={\left({\left(2\right)}^{3\left(-2\right)}\right)}^{\frac{1}{3}} \\ ={\left({\left(2\right)}^{-6}\right)}^{\frac{1}{3}} \\ ={\left(2\right)}^{-6\times \frac{1}{3}} \\ ={\left(2\right)}^{-2} \\ \mathrm{Use}\mathrm{the}\mathrm{rule}\mathrm{of}\mathrm{exponent}{\mathrm{a}}^{-\mathrm{n}}=\frac{1}{{\mathrm{a}}^{\mathrm{n}}} \\ \sqrt[3]{8^{-2}}=\frac{1}{2^2} \\ =\frac{\mathbf{1}}{\mathbf{4}} \end{gathered}$$