Question

Evaluate
(i) ${\left(125\right)}^{\frac{1}{3}}$

(ii) ${\left(64\right)}^{\frac{1}{6}}$

(iii) ${\left(25\right)}^{\frac{3}{2}}$

(iv) ${\left(81\right)}^{\frac{3}{4}}$

(v) ${\left(64\right)}^{-\frac{1}{2}}$

(vi) ${\left(8\right)}^{-\frac{1}{3}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(125\right)}^{\frac{1}{3}}={\left(5^3\right)}^{\frac{1}{3}}=5^{3\times \frac{1}{3}}=5 \\ \left(\mathrm{ii}\right){\left(64\right)}^{\frac{1}{6}}={\left(2^6\right)}^{\frac{1}{6}}=2^{6\times \frac{1}{6}}=2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right){\left(25\right)}^{\frac{3}{2}}=5^{2\times \frac{3}{2}}=5^3=125\left[(a^m{)}^n=a^{mn}\right] \\ \left(\mathrm{iv}\right){\left(81\right)}^{\frac{3}{4}}={\left(3^4\right)}^{\frac{3}{4}}=3^{4\times \frac{3}{4}}=3^3=27 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right){\left(64\right)}^{\frac{-1}{2}}={\left(8^2\right)}^{\frac{-1}{2}}=8^{2\times \frac{-1}{2}}=8^{-1}=\frac{1}{8} \\ \left(\mathrm{vi}\right){\left(8\right)}^{\frac{-1}{3}}={\left(2^3\right)}^{\frac{-1}{3}}=2^{3\times \frac{-1}{3}}=2^{-1}=\frac{1}{2} \end{gathered}$$