Evaluate the following :

$(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} . \Big(\dfrac{1}{2}\Big)^{-1} . (2)^0$

Given,

$(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} . \Big(\dfrac{1}{2}\Big)^{-1} . (2)^0$

Simplifying the expression :

$\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} \times \Big(\dfrac{1}{2}\Big)^{-1} \times (2)^0 \\[1em] \Rightarrow [(3)^4]^{\dfrac{3}{4}} - (32)^{\dfrac{2}{5}} + [(2)^3]^{\dfrac{1}{3}} \times (2) \times 1 \\[1em] \Rightarrow (3)^3 - [(2)^5]^{\dfrac{2}{5}} + 2^1 \times 2 \\[1em] \Rightarrow 27 - (2)^2 + 4 \\[1em] \Rightarrow 31 - 4 \\[1em] \Rightarrow 27.$

Hence, $(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} \times \Big(\dfrac{1}{2}\Big)^{-1} \times (2)^0 = 27$.