9. Determine the value of each expression: \( \begin{array}{llll}\text { (a) } 27^{\frac{1}{3}} & \text { (b) } 32^{\frac{3}{5}} & \text { (c) } 49^{-\frac{1}{2}} & \text { (d) }\left(\frac{16}{81}\right)^{-\frac{3}{4}}\end{array} \)

Let's evaluate each expression step by step: ### (a) \( 27^{\frac{1}{3}} \) This represents the cube root of 27. - **Calculation:** \[ 27 = 3^3 \implies 27^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = 3^{3 \times \frac{1}{3}} = 3^1 = 3 \] - **Answer:** \( 3 \) --- ### (b) \( 32^{\frac{3}{5}} \) This can be interpreted as the 5th root of 32 raised to the 3rd power. - **Express 32 as a power of 2:** \[ 32 = 2^5 \] - **Calculation:** \[ 32^{\frac{3}{5}} = (2^5)^{\frac{3}{5}} = 2^{5 \times \frac{3}{5}} = 2^3 = 8 \] - **Answer:** \( 8 \) --- ### (c) \( 49^{-\frac{1}{2}} \) A negative exponent indicates a reciprocal, and the exponent \(\frac{1}{2}\) represents a square root. - **Express 49 as a power of 7:** \[ 49 = 7^2 \] - **Calculation:** \[ 49^{-\frac{1}{2}} = (7^2)^{-\frac{1}{2}} = 7^{2 \times -\frac{1}{2}} = 7^{-1} = \frac{1}{7} \] - **Answer:** \( \frac{1}{7} \) --- ### (d) \( \left(\frac{16}{81}\right)^{-\frac{3}{4}} \) A negative exponent indicates taking the reciprocal, and the exponent \(\frac{3}{4}\) involves roots and powers. - **Express 16 and 81 as powers:** \[ 16 = 2^4 \quad \text{and} \quad 81 = 3^4 \] - **Rewrite the expression:** \[ \left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left(\frac{2^4}{3^4}\right)^{-\frac{3}{4}} = \left(\frac{2}{3}\right)^{4 \times -\frac{3}{4}} = \left(\frac{2}{3}\right)^{-3} = \left(\frac{3}{2}\right)^3 = \frac{27}{8} \] - **Answer:** \( \frac{27}{8} \) --- ### **Summary of Answers:** - **(a)** \( 3 \) - **(b)** \( 8 \) - **(c)** \( \frac{1}{7} \) - **(d)** \( \frac{27}{8} \)