Simplify the following:

$(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1}$

Given,

$\Rightarrow (27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} \\[1em] \Rightarrow (3^3)^{\dfrac{4}{3}} + (2^5)^{\dfrac{8}{10}} + \Big(\dfrac{1}{0.8}\Big)^{1} \\[1em] \Rightarrow (3)^{3 \times \dfrac{4}{3}} + (2)^{5 \times \dfrac{8}{10}} + \Big(\dfrac{10}{8}\Big) \\[1em] \Rightarrow (3)^4 + 2^4 + \dfrac{5}{4} \\[1em] \Rightarrow 81 + 16 + \dfrac{5}{4} \\[1em] \Rightarrow 97 + \dfrac{5}{4} \\[1em] \Rightarrow \dfrac{388 + 5}{4} \\[1em] \Rightarrow \dfrac{393}{4} \\[1em] \Rightarrow 98.25$

Hence, $(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1}$ = 98.25