Simplify and express each of the following as a rational number :

(i) $\Big(\dfrac{6}{5}\Big)^3 \times \Big(\dfrac{5}{2}\Big)^2$

(ii) $\Big(\dfrac{3}{4}\Big)^2 \times \Big(\dfrac{-1}{2}\Big)^5 \times 2^3$

(iii) $\Big(\dfrac{5}{4}\Big)^2 \times \Big(\dfrac{2}{3}\Big)^2 \times \Big(\dfrac{-3}{5}\Big)^3$

(iv) $\Big(\dfrac{-3}{4}\Big)^3 \times \Big(\dfrac{-5}{2}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^5$

(v) $\Big(\dfrac{7}{11}\Big)^6 ÷ \Big(\dfrac{7}{11}\Big)^3$

(vi) $\Big(\dfrac{-4}{3}\Big)^8 ÷ \Big(\dfrac{-4}{3}\Big)^{12}$

(i) We have:

$\phantom{=} \Big(\dfrac{6}{5}\Big)^3 \times \Big(\dfrac{5}{2}\Big)^2 \\[1em] = \dfrac{6^3}{5^3} \times \dfrac{5^2}{2^2} \\[1em] = \dfrac{6^3 \times 5^2}{5^3 \times 2^2} \\[1em] = \dfrac{6^3 \times 5^{2-3}}{2^2}\quad \text{[Applying exponent law]} \\[1em] = \dfrac{6^3 \times 5^{-1}}{2^2} \\[1em] = \dfrac{6 \times 6 \times 6}{5 \times 2 \times 2} \quad {[5^{-1} = \dfrac{1}{5}]} \\[1em] = \dfrac{216}{20} \\[1em] = \dfrac{54}{5} = 10\dfrac{4}{5}$

Hence, the answer is $10\dfrac{4}{5}$

(ii) We have:

$\phantom{=} \Big(\dfrac{3}{4}\Big)^2 \times \Big(\dfrac{-1}{2}\Big)^5 \times 2^3 \\[1em] = \dfrac{3^2}{4^2} \times \dfrac{(-1)^5}{2^5} \times 2^3 \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^3}{4^2 \times 2^5} \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^{3-5}}{4^2} \quad \text{[Applying exponent law]} \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^{-2}}{4^2} \\[1em] = \dfrac{(3 \times 3) \times (-1)}{4 \times (4 \times 4)} \quad [2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}] \\[1em] = \dfrac{-9}{64}$

Hence, the answer is $\dfrac{-9}{64}$

(iii) We have:

$\phantom{=} \Big(\dfrac{5}{4}\Big)^2 \times \Big(\dfrac{2}{3}\Big)^2 \times \Big(\dfrac{-3}{5}\Big)^3 \\[1em] \dfrac{5^2}{4^2} \times \dfrac{2^2}{3^2} \times \dfrac{(-3)^3}{5^3} \\[1em] = \dfrac{5^2 \times 2^2 \times (-3)^3}{4^2 \times 3^2 \times 5^3} \\[1em] = \dfrac{2^2 \times (-3)^{3-2} \times 5^{2-3}}{4^2} \quad \text{[Applying exponent law]} \\[1em] = \dfrac{2^2 \times (-3)^1 \times 5^{-1}}{4^2} \\[1em] = \dfrac{(2 \times 2) \times (-3)}{(4 \times 4) \times 5} \quad[5^{-1} = \dfrac{1}{5}] \\[1em] = \dfrac{-12}{80} = \dfrac{-3}{20}$

Hence, the answer is $\dfrac{-3}{20}$

(iv) We have:

$\phantom{=} \Big(\dfrac{-3}{4}\Big)^3 \times \Big(\dfrac{-5}{2}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^5 \\[1em] = \dfrac{(-3)^3}{4^3} \times \dfrac{(-5)^3}{2^3} \times \dfrac{2^5}{3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^5}{4^3 \times 2^3 \times 3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^{5-3}}{4^3 \times 3^5} \quad \text{[Applying exponent law]} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^2}{4^3 \times 3^5} \\[1em] = \dfrac{(-3 \times -3 \times -3) \times (-5 \times -5 \times -5) \times (2 \times 2)}{(4 \times 4 \times 4) \times (3 \times 3 \times 3 \times 3 \times 3)} \\[1em] = \dfrac{(-27) \times -125 \times 4}{64 \times 243} \\[1em] = \dfrac{-1 \times -125 \times 1}{16 \times 9} \quad \text{[Dividing 27 and 243 by 27, 4 and 64 by 4]} \\[1em] = \dfrac{125}{144} \\[1em]$

Hence, the answer is $\dfrac{125}{144}$

(v) We have:

$\phantom{=} \Big(\dfrac{7}{11}\Big)^6 ÷ \Big(\dfrac{7}{11}\Big)^3 \\[1em] = \Big(\dfrac{7}{11}\Big)^{6-3} \quad \text{[By division rule]} \\[1em] = \Big(\dfrac{7}{11}\Big)^3 \\[1em] = \dfrac{7^3}{11^3} \\[1em] = \dfrac{7 \times 7 \times 7}{11 \times 11 \times 11} \\[1em] = \dfrac{343}{1331}$

Hence, the answer is $\dfrac{343}{1331}$

(vi) We have:

$\phantom{=} \Big(\dfrac{-4}{3}\Big)^8 ÷ \Big(\dfrac{-4}{3}\Big)^{12} \\[1em] = \Big(\dfrac{-4}{3}\Big)^{8-12} \quad \text{[By division rule]} \\[1em] = \Big(\dfrac{-4}{3}\Big)^{-4} \\[1em] = \Big(\dfrac{3}{-4}\Big)^{4} \quad \text{[By reciprocal rule]} \\[1em] = \dfrac{3 \times 3 \times 3 \times 3}{(-4) \times (-4) \times (-4) \times (-4)} \\[1em] = \dfrac{81}{256}$

Hence, the answer is $\dfrac{81}{256}$