Simplify the following

(i) $\left(1^{3}+2^{3}+3^{3}\right)^{\frac{1}{2}}=(1+8+27)^{\frac{1}{2}}$

$=(36)^{\frac{1}{2}}=\left(6^{2}\right)^{\frac{1}{2}}=6^{2 \times \frac{1}{2}}=6$            $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

(ii) $\left(\frac{3}{5}\right)^{4}\left(\frac{8}{5}\right)^{-12}\left(\frac{32}{5}\right)^{6}=\frac{3^{4}}{5^{4}} \times\left(\frac{5}{2^{3}}\right)^{12} \times\left(\frac{2^{5}}{5}\right)^{6}\left(\because a^{-1}=\frac{1}{a}\right)$                  

$=\frac{3^{4}}{5^{4}} \times \frac{5^{12}}{2^{36}} \times \frac{2^{30}}{5^{6}}$            $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=\frac{3^{4} \times 5^{12-4-6}}{2^{36-30}}$        $\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]$

$=\frac{3^{4}}{2^{6}} \times 5^{2}=\frac{81 \times 25}{64}=\frac{2025}{64}$

(iii) $\left(\frac{1}{27}\right)^{-2 / 3}=\left(\frac{1}{3^{3}}\right)^{-2 / 3}=\left(3^{-3}\right)^{-2 / 3}$                   $\left[\because \frac{1}{a}=a^{-1}\right]$

$=3^{-3 \cdot x-\frac{2}{3}}=3^{2}=9$             $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

(iv) $\left.\left.\left[(625)^{-\frac{1}{2}}\right)^{\frac{-1}{4}}\right]^{2}=\left[\left(25^{2}\right)^{-\frac{1}{2}}\right)^{-\frac{1}{4}}\right]^{2}=\left(25^{-1}\right)^{-\frac{1}{4} \times 2}=\left[\left(5^{2}\right)^{-1}\right]^{-\frac{1}{4} \times 2}$              $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=5^{-2 \times-\frac{1}{4} \times 2}=5^{1}=5$

(v) $\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}=\frac{\left(3^{2}\right)^{\frac{1}{3}} \times\left(3^{3}\right)^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}$                               $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=\frac{3^{\frac{2}{3}} \times 3^{-\frac{3}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}=\frac{3^{\frac{2}{3}-\frac{3}{2}}}{3^{\frac{1}{6}-\frac{2}{3}}}$                   $\left[\because a^{m} \times a^{n}=a^{m+n}\right]$

$=\frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}=3^{-\frac{5}{6}+\frac{3}{6}}=3^{-\frac{2}{6}}=3^{-\frac{1}{3}}$                 $\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]$

(vi) $64^{-\frac{1}{3}}\left[64^{\frac{1}{3}}-64^{\frac{2}{3}}\right]=\left(4^{3}\right)^{-\frac{1}{3}}\left[\left(4^{3}\right)^{\frac{1}{3}}-\left(4^{3}\right)^{\frac{2}{3}}\right]$

$=4^{3 \times-\frac{1}{3}}\left(4^{3 \times \frac{1}{3}}-4^{3 \times \frac{2}{3}}\right)=4^{-1}\left(4-4^{2}\right)$                      $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=\frac{1}{4}(4-16)=-\frac{12}{4}=-3$

(vii) $\frac{8^{\frac{1}{3}} \times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}=\frac{\left(2^{3}\right)^{\frac{1}{3}} \times\left(2^{4}\right)^{\frac{1}{3}}}{\left(2^{5}\right)^{-\frac{1}{3}}}=\frac{2^{3 \times \frac{1}{3}} \times 2^{4 \times \frac{1}{3}}}{2^{5 \times-\frac{1}{3}}}$                    $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$

$=\frac{2^{\frac{3}{3}+\frac{4}{3}}}{2^{-\frac{5}{3}}}=\frac{2^{\frac{7}{3}}}{2^{-\frac{5}{3}}}=2^{\frac{7}{3}+\frac{5}{3}}=2^{\frac{12}{3}}=2^{4}=16$                 $\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]$