Prove that

Solution:

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

$\mathrm{LHS}=\left[8^{-\frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{-\frac{5}{4}}\right] \div\left[32^{-\frac{2}{5}} \times 125^{-\frac{5}{6}}\right]$

$=\left[\left[(2)^{3}\right]^{-\frac{2}{3}} \times 2^{\frac{1}{2}} \times\left[(5)^{2}\right]^{-\frac{5}{4}}\right] \div\left[\left[(2)^{5}\right]^{-\frac{2}{5}} \times\left[(5)^{3}\right]^{-\frac{5}{6}}\right]$

$=\left[2^{-2} \times 2^{\frac{1}{2}} \times 5^{-\frac{5}{2}}\right] \div\left[2^{-2} \times 5^{-\frac{5}{2}}\right]$

$=\frac{2^{-2} \times 2^{\frac{1}{2}} \times 5^{-\frac{3}{2}}}{2^{-2} \times 5^{-\frac{5}{2}}}$

$=2^{\frac{1}{2}}$

$=\sqrt{2}$

$=\mathrm{RHS}$

$\therefore\left[8^{-\frac{2}{3}} \times 2^{\frac{1}{2}} \times 25^{-\frac{5}{4}}\right] \div\left[32^{-\frac{2}{5}} \times 125^{-\frac{5}{6}}\right]=\sqrt{2}$

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

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

$=\left[\left(\frac{4}{5}\right)^{3}\right]^{-\frac{2}{3}}+\frac{1}{\left[\left(\frac{4}{5}\right)^{4}\right]^{\frac{1}{4}}}+\frac{\sqrt{5 \times 5}}{\sqrt[3]{4 \times 4 \times 4}}$

$=\left(\frac{4}{5}\right)^{-2}+\frac{1}{\left(\frac{4}{5}\right)}+\frac{5}{4}$

$=\left(\frac{5}{4}\right)^{2}+\frac{5}{4}+\frac{5}{4}$

$=\frac{25}{16}+\frac{5}{4}+\frac{5}{4}$

$=\frac{25+20+20}{16}$

$=\frac{65}{16}$

$=\mathrm{RHS}$

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

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

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

$=\left\{7\left[\left((3)^{4}\right)^{\frac{1}{4}}+\left((4)^{4}\right)^{\frac{1}{4}}\right]^{\frac{1}{4}}\right\}^{4}$

$=\left\{7[3+4]^{\frac{1}{4}}\right\}^{4}$

$=\left\{7^{1} \times(7)^{\frac{1}{4}}\right\}^{4}$

$=\left\{7^{1+\frac{1}{4}}\right\}^{4}$

$=\left\{7^{\frac{5}{4}}\right\}^{4}$

$=7^{5}$

$=16807$

$=\mathrm{RHS}$

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