What is the value of \(125^{\frac{2}{3}}\times32^{-\frac{2}{5}}\div64^{-\frac{1}{3}}\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 25
Write each base as a power of a small number. Then \(a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}\), and a negative power means the reciprocal: \(a^{-m}=\frac{1}{a^{m}}\).
\(125^{\frac{2}{3}}=\left(\sqrt[3]{125}\right)^{2}=5^{2}=25\)
\(\displaystyle32^{-\frac{2}{5}}=\frac{1}{\left(\sqrt[5]{32}\right)^{2}}=\frac{1}{2^{2}}=\frac{1}{4}\)
\(\displaystyle64^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{64}}=\frac{1}{4}\)
\(\displaystyle25\times\frac{1}{4}\div\frac{1}{4}=25\times\frac{1}{4}\times4=25\)