Simplify the following:
(8125)−13\Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}}(1258)−31
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Given,
⇒(8125)−13=(1258)13=(5323)13=53×1323×13=52=212.\Rightarrow \Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} = \Big(\dfrac{125}{8}\Big)^{\dfrac{1}{3}} \\[1em] = \Big(\dfrac{5^3}{2^3}\Big)^{\dfrac{1}{3}} = \dfrac{5^{3 \times \dfrac{1}{3}}}{2^{3 \times \dfrac{1}{3}}}\\[1em] = \dfrac{5}{2} = 2\dfrac{1}{2}.⇒(1258)−31=(8125)31=(2353)31=23×3153×31=25=221.
Hence, (8125)−13=212.\Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} = 2\dfrac{1}{2}.(1258)−31=221.
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