x*x*x is equal to 2 Prove and Solved

to solve the equation: x*x*x is equal to 2 \[ x^3 = 2 \] To find \( x \), take the cube root of both sides: \[ x = \sqrt[3]{2} \] Approximating the value: \[ x \approx 1.2599 \] If you need the exact cube roots in complex form, they are: \[ x = \sqrt[3]{2}, \quad x = \sqrt[3]{2} \cdot e^{2\pi i / 3}, \quad x = \sqrt[3]{2} \cdot e^{4\pi i / 3} \] Which simplifies to: \[ x_1 \approx 1.2599 \] \[ x_2 \approx -0.6299 + i \cdot 1.0911 \] \[ x_3 \approx -0.6299 – i \cdot 1.0911 \] Let me know if you need further explanation! 😊

To **prove** that \( x^3 = 2 \), you need to show that there exists a value of \( x \) that satisfies this equation. We can express \( x \) explicitly as: \[ x = \sqrt[3]{2} \] ### Step 1: Cube Both Sides Cubing both sides, \[ (\sqrt[3]{2})^3 = 2 \] which simplifies to: \[ x^3 = 2 \] This proves that \( x = \sqrt[3]{2} \) is a valid solution. ### Step 2: Consider the Complex Cube Roots Using Euler’s formula, the general solution for the cube roots of 2 is: \[ x_k = \sqrt[3]{2} e^{2\pi i k / 3}, \quad k = 0, 1, 2 \] These correspond to: \[ x_0 = \sqrt[3]{2} \] \[ x_1 = \sqrt[3]{2} \cdot e^{2\pi i / 3} = -\frac{\sqrt[3]{2}}{2} + i\frac{\sqrt[3]{2} \sqrt{3}}{2} \] \[ x_2 = \sqrt[3]{2} \cdot e^{4\pi i / 3} = -\frac{\sqrt[3]{2}}{2} – i\frac{\sqrt[3]{2} \sqrt{3}}{2} \] Each of these values satisfies \( x^3 = 2 \), proving that there are three valid solutions in the complex plane.