Let's break this problem into **two steps** as requested: --- ### **Step 1: Write Down the Equations and the Newton's Method Update Formula** Given the equations: - \( f(u, v) = u^3 - v = 0 \) - \( g(u, v) = v^3 - u = 0 \) Let \(\mathbf{x}_0 = (u_0, v_0)\) be the starting point. For Newton's method in two variables, the update rule is: \[ \mathbf{x}_1 = \mathbf{x}_0 - J^{-1} F(\mathbf{x}_0) \] where \(F(\mathbf{x}_0) = \begin{bmatrix} f(u_0, v_0) \\ g(u_0, v_0) \end{bmatrix}\) and \(J\) is the Jacobian matrix of partial derivatives. **Explanation:** This sets up the system and the Newton's method formula we will use. Next, we compute the Jacobian and carry out the update step. --- ### **Step 2: Compute the Jacobian, its Inverse, and Apply Newton's Step** **Jacobian Matrix:** \[ J = \begin{bmatrix} \frac{\partial f}{\partial u} & \frac{\partial f}{\partial v} \\ \frac{\partial g}{\partial u} & \frac{\partial g}{\partial v} \end{bmatrix} = \begin{bmatrix} 3u^2 & -1 \\ -1 & 3v^2 \end{bmatrix} \] At \((u_0, v_0)\), the Jacobian is: \[ J_0 = \begin{bmatrix} 3u_0^2 & -1 \\ -1 & 3v_0^2 \end{bmatrix} \] **Function Vector:** \[ F(\mathbf{x}_0) = \begin{bmatrix} u_0^3 - v_0 \\ v_0^3 - u_0 \end{bmatrix} \] **Inverse of 2x2 Jacobian:** \[ J_0^{-1} = \frac{1}{\det J_0} \begin{bmatrix} 3v_0^2 & 1 \\ 1 & 3u_0^2 \end{bmatrix} \] where \(\det J_0 = 3u_0^2 \cdot 3v_0^2 - (-1) \cdot (-1) = 9u_0^2 v_0^2 - 1\). **Newton's Step:** \[ \begin{bmatrix} u_1 \\ v_1 \end{bmatrix} = \begin{bmatrix} u_0 \\ v_0 \end{bmatrix} - J_0^{-1} \begin{bmatrix} u_0^3 - v_0 \\ v_0^3 - u_0 \end{bmatrix} \] Plugging in the inverse: \[ \begin{bmatrix} u_1 \\ v_1 \end{bmatrix} = \begin{bmatrix} u_0 \\ v_0 \end{bmatrix} - \frac{1}{9u_0^2 v_0^2 - 1} \begin{bmatrix} 3v_0^2 & 1 \\ 1 & 3u_0^2 \end{bmatrix} \begin{bmatrix} u_0^3 - v_0 \\ v_0^3 - u_0 \end{bmatrix} \] **Explanation:** In this step, we calculated the Jacobian, its determinant, and its inverse, then applied Newton's method to get the explicit update for \((u_1, v_1)\) in terms of \((u_0, v_0)\). --- ## **Final Answer** \[ \boxed{ \begin{aligned} u_1 &= u_0 - \frac{1}{9u_0^2 v_0^2 - 1} \left[ 3v_0^2(u_0^3 - v_0) + (v_0^3 - u_0) \right] \\ v_1 &= v_0 - \frac{1}{9u_0^2 v_0^2 - 1} \left[ (u_0^3 - v_0) + 3u_0^2(v_0^3 - u_0) \right] \end{aligned} } \] This gives the next point \((u_1, v_1)\) from the starting point \((u_0, v_0)\) using Newton's method for the given system.