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We are tasked with approximating the value of \( \sqrt[3]{40} \) using Newton's method.

### Newton's Method for Cube Roots
To apply Newton's method to this problem, we want to solve the equation:
\[
f(x) = x^3 - 40 = 0
\]
The corresponding derivative is:
\[
f'(x) = 3x^2
\]
Newton's method uses the iteration formula:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
Substituting for \( f(x) \) and \( f'(x) \), the iteration becomes:
\[
x_{n+1} = x_n - \frac{x_n^3 - 40}{3x_n^2}
\]
This simplifies to:
\[
x_{n+1} = \frac{2x_n + \frac{40}{x_n^2}}{3}
\]

### Step-by-Step Solution:
Given that \( x_1 = 3 \), we can now find \( x_2 \) and \( x_3 \).

1. **First Iteration to Find \( x_2 \):**
\[
x_2 = \frac{2(3) + \frac{40}{3^2}}{3} = \frac{6 + \frac{40}{9}}{3} = \frac{6 + 4.44}{3} = \frac{10.44}{3} = 3.48
\]

2. **Second Iteration to Find \( x_3 \):**
\[
x_3 = \frac{2(3.48) + \frac{40}{3.48^2}}{3} = \frac{6.96 + \frac{40}{12.11}}{3} = \frac{6.96 + 3.3}{3} = \frac{10.26}{3} = 3.42
\]

Thus, the second approximation \( x_2 \) is approximately **3.48**, and the third approximation \( x_3 \) is approximately **3.42**.

Let me know if you'd like more details or further clarifications!

Here are 5 related questions for further exploration:
1. How does Newton's method behave for different initial approximations?
2. Why does Newton's method converge to the cube root in this case?
3. What is the role of the derivative in the Newton iteration formula?
4. How many iterations are typically required for sufficient accuracy in Newton's method?
5. How would Newton's method change if applied to different root-finding problems, like square roots?

**Tip:** The closer your initial guess is to the actual root, the faster Newton's method converges.

Use Newton's method to approximate the value of cube root of 40. Let x1 = 3 be the initial approximation. Find the second approximation x2 and the third approximation x3.

This page explains how to use Newton's method to approximate the cube root of 40, starting with an initial guess of x1 = 3. The method involves solving f(x) = x^3 - 40 = 0 using the iteration formula. Follow the detailed steps to find the second and third approximations, x2 and x3.

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