Prove: \[ \left(\frac{64}{125}\right)^{-2/3} + \frac{1}{\left(\frac{256}{625}\right)^{1/4}} + \left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0 = \frac{61}{16} \]
Proof
\[ \left(\frac{64}{125}\right)^{-2/3} = \left(\frac{4^3}{5^3}\right)^{-2/3} = \left(\frac{4}{5}\right)^{-2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]
\[ \frac{1}{\left(\frac{256}{625}\right)^{1/4}} = \frac{1}{\left(\frac{4^4}{5^4}\right)^{1/4}} = \frac{1}{\frac{4}{5}} = \frac{5}{4} \]
\[ \left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0 = 1 \]
\[ = \frac{25}{16} + \frac{5}{4} + 1 \]
\[ = \frac{25}{16} + \frac{20}{16} + \frac{16}{16} \]
\[ = \frac{61}{16} \]