Simplify the following:
(64125)−23÷1(256625)14+(25643)0\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0(12564)−32÷(625256)411+(36425)0
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Given,
⇒(64125)−23÷1(256625)14+(25643)0=(12564)23×(256625)14+1=[(54)3]23×[(45)4]14+1=(54)2×45+1=54+1=94=214.\Rightarrow \Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 \\[1em] = \Big(\dfrac{125}{64}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}} + 1 \\[1em] = \Big[\Big(\dfrac{5}{4}\Big)^3\Big]^{\dfrac{2}{3}} \times \Big[\Big(\dfrac{4}{5}\Big)^4\Big]^{\dfrac{1}{4}} + 1 \\[1em] = \Big(\dfrac{5}{4}\Big)^2 \times \dfrac{4}{5} + 1 \\[1em] = \dfrac{5}{4} + 1 \\[1em] = \dfrac{9}{4} = 2\dfrac{1}{4}.⇒(12564)−32÷(625256)411+(36425)0=(64125)32×(625256)41+1=[(45)3]32×[(54)4]41+1=(45)2×54+1=45+1=49=241.
Hence, (64125)−23÷1(256625)14+(25643)0=214\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 = 2\dfrac{1}{4}(12564)−32÷(625256)411+(36425)0=241
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