Simplify:

(i) $\left(2^{\smash{\frac{2}{3}}} \right)$. $\left(2^{\smash{\frac{1}{5}}} \right)$

(ii) $\Big(\dfrac{1}{3^3}\Big)^7$

(iii) $\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}$

(iv) $\left(7^{\smash{\frac{1}{2}}} \right)$. $\left(8^{\smash{\frac{1}{2}}} \right)$

(i) $\left(2^{\smash{\frac{2}{3}}} \right)$. $\left(2^{\smash{\frac{1}{5}}} \right)$

= $\left(2^{\smash{\frac{2}{3} + \frac{1}{5}}} \right)$

= $\left((2)^{\smash{\frac{10+3}{15}}} \right)$

= $\left((2)^{\smash{\frac{13}{15}}} \right)$

Hence, $\left(2^{\smash{\frac{2}{3}}} \right)$. $\left(2^{\smash{\frac{1}{5}}} \right)$ = $\left((2)^{\smash{\frac{13}{15}}} \right)$

(ii) $\Big(\dfrac{1}{3^3}\Big)^7$

$= \Big(\dfrac{1}{3^3}\Big)^7 \\[1em] = \Big(\dfrac{1^7}{3^{3 \times 7}}\Big) \\[1em] = \Big(\dfrac{1}{3^{21}}\Big)$

Hence, $\Big(\dfrac{1}{3^3}\Big)^7 = \Big(\dfrac{1}{3^{21}}\Big)$

(iii) $\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}$

= $\left(11^{\smash{\frac{1}{2} - \frac{1}{4}}} \right)$

= $\left(11^{\smash{\frac{2-1}{4}}} \right)$

= $\left(11^{\smash{\frac{1}{4}}} \right)$

Hence, $\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}$ = $\left(11^{\smash{\frac{1}{4}}} \right)$

(iv) $\left(7^{\smash{\frac{1}{2}}} \right)$. $\left(8^{\smash{\frac{1}{2}}} \right)$

$= (7 \times 8)^{\dfrac{1}{2}}\quad[\because \text{a}^m \times \text{b}^m = (\text{ab})^m] \\[1em] = 56^{\frac{1}{2}}$

Hence, $\left(7^{\smash{\frac{1}{2}}} \right).\left(8^{\smash{\frac{1}{2}}} \right) = 56^{\frac{1}{2}}$