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Mathematics

Evaluate :

(64)23−1253−12−5+(27)−23×(259)−12(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}}

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Answer

Simplifying the expression :

⇒(64)23−1253−12−5+(27)−23×(259)−12=(26)23−(53)13−25+(33)−23×[(53)2]−12=(2)6×23−(5)3×13−32+(3)3×−23×(53)2×−12=24−51−32+3−2×(53)−1=16−5−32+132×35=−21+19×35=−21+345=−21+115=−315+115=−31415=−201415.\Rightarrow (64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} \\[1em] = (2^6)^{\dfrac{2}{3}} - (5^3)^{\dfrac{1}{3}} - 2^5 + (3^3)^{-\dfrac{2}{3}} \times \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-\dfrac{1}{2}}\\[1em] = (2)^{6 \times \dfrac{2}{3}} - (5)^{3 \times \dfrac{1}{3}} - 32 + (3)^{3 \times -\dfrac{2}{3}} \times \Big(\dfrac{5}{3}\Big)^{2 \times -\dfrac{1}{2}} \\[1em] = 2^4 - 5^1 - 32 + 3^{-2} \times \Big(\dfrac{5}{3}\Big)^{-1} \\[1em] = 16 - 5 - 32 + \dfrac{1}{3^2} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{1}{9} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{3}{45} \\[1em] = -21 + \dfrac{1}{15} \\[1em] = \dfrac{-315 + 1}{15} \\[1em] = \dfrac{-314}{15} \\[1em] = -20\dfrac{14}{15}.

Hence, (64)23−1253−12−5+(27)−23×(259)−12=−201415.(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} = -20\dfrac{14}{15}.

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