\( 4 x^{3}=27 \quad \) b, \( 9 x^{3}=72 \quad \) c, \( \frac{1}{3} x^{5}=81 \) d \( 2 x^{3}-4=2 x 6 \)

Sure, let's solve each of the given equations step by step. --- ### **Problem b** \[ 4x^{3} = 27 \] **Solution:** 1. **Isolate \( x^3 \):** \[ x^{3} = \frac{27}{4} \] 2. **Take the cube root of both sides:** \[ x = \sqrt[3]{\frac{27}{4}} = \frac{\sqrt[3]{27}}{\sqrt[3]{4}} = \frac{3}{\sqrt[3]{4}} \] Alternatively, you can write: \[ x = 3 \cdot 4^{-\frac{1}{3}} \] --- ### **Problem c** \[ 9x^{3} = 72 \] **Solution:** 1. **Isolate \( x^3 \):** \[ x^{3} = \frac{72}{9} = 8 \] 2. **Take the cube root of both sides:** \[ x = \sqrt[3]{8} = 2 \] --- ### **Problem d** This problem appears to include two separate equations: 1. \(\frac{1}{3}x^{5} = 81\) 2. \(2x^{3} - 4 = 2x + 6\) Let's solve each one individually. --- #### **Equation 1:** \[ \frac{1}{3}x^{5} = 81 \] **Solution:** 1. **Multiply both sides by 3 to isolate \( x^5 \):** \[ x^{5} = 81 \times 3 = 243 \] 2. **Take the fifth root of both sides:** \[ x = \sqrt[5]{243} \] Since \( 243 = 3^5 \): \[ x = 3 \] --- #### **Equation 2:** \[ 2x^{3} - 4 = 2x + 6 \] **Solution:** 1. **Rearrange the equation to set it to zero:** \[ 2x^{3} - 2x - 10 = 0 \] 2. **Simplify by dividing the entire equation by 2:** \[ x^{3} - x - 5 = 0 \] 3. **Solve for \( x \) using numerical methods (since it's a cubic equation without easily factorable roots):** - **Initial Guess:** Let’s try \( x = 2 \): \[ f(2) = 2^3 - 2 - 5 = 8 - 2 - 5 = 1 \quad (\text{Too high}) \] - **Next Guess:** Try \( x = 1.9 \): \[ f(1.9) = 1.9^3 - 1.9 - 5 \approx 6.859 - 1.9 - 5 = -0.041 \] - **Refinement:** Using methods like Newton-Raphson, we find: \[ x \approx 1.903 \] **Thus, the approximate real solution is:** \[ x \approx 1.903 \] --- ### **Summary of Solutions** - **Problem b:** \( x = \dfrac{3}{\sqrt[3]{4}} \) or \( x = 3 \cdot 4^{-\frac{1}{3}} \) - **Problem c:** \( x = 2 \) - **Problem d:** - First equation: \( x = 3 \) - Second equation: \( x \approx 1.903 \) If you need further clarification or assistance with any of these problems, feel free to ask!