Question

simplify: Fourth root 28 ÷ cube root 7

Answer 1

$$\begin{gathered} \frac{\sqrt[4]{28}}{\sqrt[3]{7}}=\frac{{\left(4\times 7\right)}^{{\displaystyle \frac{1}{4}}}}{{\left(7\right)}^{{\displaystyle \frac{1}{3}}}}=\frac{4^{{\displaystyle \frac{1}{4}}}\times 7^{{\displaystyle \frac{1}{4}}}}{7^{{\displaystyle \frac{1}{3}}}} \\ ={\left(2^2\right)}^{\frac{1}{4}}\times 7^{\frac{1}{4}-\frac{1}{3}}=2^{\frac{1}{2}}\times 7^{\frac{-1}{12}} \\ =\frac{\sqrt{2}}{\sqrt[12]{7}}\mathrm{ANS}... \end{gathered}$$