(b) \( \left(\frac{25}{16}\right)^{\frac{-1}{2}}\left(\frac{125}{64}\right)^{\frac{1}{3}}=\left(\frac{8}{27}\right)^{\frac{-1}{3}} \)

Let’s break this down step by step in a fun and easy way! First, for \( \left(\frac{25}{16}\right)^{\frac{-1}{2}} \), the negative exponent tells us to take the reciprocal of the base. So we flip it over and get \( \left(\frac{16}{25}\right)^{\frac{1}{2}} \). Now taking the square root, we have: \[ \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \] Next, we look at \( \left(\frac{125}{64}\right)^{\frac{1}{3}} \). This means we take the cube root: \[ \sqrt[3]{\frac{125}{64}} = \frac{\sqrt[3]{125}}{\sqrt[3]{64}} = \frac{5}{4} \] So now we can combine what we’ve got: \[ \left(\frac{25}{16}\right)^{\frac{-1}{2}}\left(\frac{125}{64}\right)^{\frac{1}{3}} = \frac{4}{5} \times \frac{5}{4} = 1 \] Now, let’s tackle \( \left(\frac{8}{27}\right)^{\frac{-1}{3}} \): This gives us \( \left(\frac{27}{8}\right)^{\frac{1}{3}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2} \). We see that \( 1 \) is not equal to \( \frac{3}{2} \), so the original equation is incorrect. Thus, \[ \left(\frac{25}{16}\right)^{\frac{-1}{2}}\left(\frac{125}{64}\right)^{\frac{1}{3}} \neq \left(\frac{8}{27}\right)^{\frac{-1}{3}} \] Overall, it’s clear we’ve made some misjudgments in assuming equivalence — mathematical challenges are a fun way to hone our skills!