Question

simplify : root25/ ³root64 + (256/625)^-1/4 + 1/(64/125)^2/3

Answer 1

Dear student
$$\begin{gathered} \frac{\sqrt{25}}{\sqrt[3]{64}}+{\left(\frac{256}{625}\right)}^{\frac{-1}{4}}+\frac{1}{{\displaystyle {\left({\displaystyle \frac{64}{125}}\right)}^{\frac{2}{3}}}} \\ =\frac{5}{4}+{\left(\frac{625}{256}\right)}^{\frac{1}{4}}+\frac{1}{{\left({\left({\displaystyle \frac{4}{5}}\right)}^3\right)}^{{\displaystyle \frac{2}{3}}}}\left[\because \sqrt[3]{64}=\sqrt[3]{4\times 4\times 4}=4\mathrm{and}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right] \\ =\frac{5}{4}+{\left[{\left(\frac{5}{4}\right)}^4\right]}^{\frac{1}{4}}+\frac{1}{{\left({\displaystyle \frac{4}{5}}\right)}^2}\left[\because {\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}\times \mathrm{n}}\right] \\ =\frac{5}{4}+\frac{5}{4}+{\left(\frac{5}{4}\right)}^2 \\ =\frac{10}{4}+\frac{25}{16} \\ =\frac{40+25}{16} \\ =\frac{65}{16} \end{gathered}$$
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