Question

Prove that
(i) $\left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right]=\sqrt{2}$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$

(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^4=16807$

Answer 1

(i) $\left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right]=\sqrt{2}$
$$\begin{gathered} \mathrm{LHS}=\left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right] \\ =\left[{\left[{\left(2\right)}^3\right]}^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {\left[{\left(5\right)}^2\right]}^{-\frac{5}{4}}\right]÷\left[{\left[{\left(2\right)}^5\right]}^{-\frac{2}{5}}\times {\left[{\left(5\right)}^3\right]}^{-\frac{5}{6}}\right] \\ =\left[2^{-2}\times 2^{\frac{1}{2}}\times 5^{-\frac{5}{2}}\right]÷\left[2^{-2}\times 5^{-\frac{5}{2}}\right] \\ =\frac{2^{-2}\times 2^{\frac{1}{2}}\times 5^{-\frac{5}{2}}}{2^{-2}\times 5^{-\frac{5}{2}}} \\ =2^{\frac{1}{2}} \\ =\sqrt{2} \\ =\mathrm{RHS} \\ \therefore \left[8^{-\frac{2}{3}}\times 2^{\frac{1}{2}}\times {25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}\times {125}^{-\frac{5}{6}}\right]=\sqrt{2} \end{gathered}$$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$
$$\begin{gathered} \mathrm{LHS}={\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}} \\ ={\left[{\left(\frac{4}{5}\right)}^3\right]}^{-\frac{2}{3}}+\frac{1}{{\left[{\left({\displaystyle \frac{4}{5}}\right)}^4\right]}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{5\times 5}}{\sqrt[3]{4\times 4\times 4}} \\ ={\left(\frac{4}{5}\right)}^{-2}+\frac{1}{\left({\displaystyle \frac{4}{5}}\right)}+\frac{5}{4} \\ ={\left(\frac{5}{4}\right)}^2+\frac{5}{4}+\frac{5}{4} \\ =\frac{25}{16}+\frac{5}{4}+\frac{5}{4} \\ =\frac{25+20+20}{16} \\ =\frac{65}{16} \\ =\mathrm{RHS} \\ \therefore {\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left({\displaystyle \frac{256}{625}}\right)}^{{\displaystyle \frac{1}{4}}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16} \end{gathered}$$

(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^4=16807$
$$\begin{gathered} \mathrm{LHS}={\left\{7{\left[{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7{\left[{\left({\left(3\right)}^4\right)}^{\frac{1}{4}}+{\left({\left(4\right)}^4\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7{\left[3+4\right]}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7^1\times {\left(7\right)}^{\frac{1}{4}}\right\}}^4 \\ ={\left\{7^{1+\frac{1}{4}}\right\}}^4 \\ ={\left\{7^{\frac{5}{4}}\right\}}^4 \\ =7^5 \\ =16807 \\ =\mathrm{RHS} \\ \therefore {\left\{7{\left[{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right]}^{\frac{1}{4}}\right\}}^4=16807 \end{gathered}$$