Evaluate :

$\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}}$

Simplifying the expression :

$\Rightarrow \Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} = \Big[\Big(\dfrac{2}{3}\Big)^4\Big]^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{7}{3}\Big)^2\Big]^{\dfrac{3}{2}} ÷ \Big[\Big(\dfrac{7}{6}\Big)^3\Big]^{\dfrac{2}{3}} \\[1em] = \Big(\dfrac{2}{3}\Big)^{4 \times -\dfrac{3}{4}} \times \Big(\dfrac{7}{3}\Big)^{2 \times \dfrac{3}{2}} ÷ \Big(\dfrac{7}{6}\Big)^{3 \times \dfrac{2}{3}} \\[1em] = \Big(\dfrac{2}{3}\Big)^{-3} \times \Big(\dfrac{7}{3}\Big)^3 ÷ \Big(\dfrac{7}{6}\Big)^2 \\[1em] = \Big(\dfrac{3}{2}\Big)^3 \times \Big(\dfrac{7}{3}\Big)^3 \times \Big(\dfrac{6}{7}\Big)^2 \\[1em] = \dfrac{3^3 \times 7^3 \times 6^2}{2^3 \times 3^3 \times 7^2} \\[1em] = \dfrac{7 \times 6^2}{2^3} \\[1em] = \dfrac{252}{8} \\[1em] = 31.5$

Hence, $\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} = 31.5$