Newton's Method: To solve the equation
\[
f(x)=0
\]
by Newton's Method we start with a good initial guess $x_{0}$ and then run the iteration
\[
x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}, \quad n=0,1,2, \ldots
\]
until we get an approximation $x_{n+1}$ that is good enough for our purposes.
Suppose.you want to compute the cube root of 4 by solving the equation
\[
x^{3}-4=0
\]

Since $1^{3}=1$ and $2^{3}=8$ Let's start with
\[
x_{0}=1.5
\]

Then
\[
\begin{array}{l}
x_{1}=\square \\
x_{2}=\square \\
x_{3}=\square \text {, and }
\end{array}
\]

To check your answer compute $x_{3}^{3}=\square$.
Enter $x_{1}, x_{2}$ and $x_{3}$ with at least 6 correct digits beyond the decimal point.

Step 1 ：Newton's Method: To solve the equation \(f(x)=0\) by Newton's Method we start with a good initial guess \(x_{0}\) and then run the iteration \(x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}, \quad n=0,1,2, \ldots\) until we get an approximation \(x_{n+1}\) that is good enough for our purposes.

Step 2 ：We want to compute the cube root of 4 by solving the equation \(x^{3}-4=0\). Since \(1^{3}=1\) and \(2^{3}=8\), we start with \(x_{0}=1.5\).

Step 3 ：Using the Newton's method, we calculate the next three iterations: \(x_{1}\), \(x_{2}\), and \(x_{3}\).

Step 4 ：The values of \(x_{1}\), \(x_{2}\), and \(x_{3}\) are approximately 1.6666666666666667, 1.5873015873015872, and 1.587401052148227 respectively.

Step 5 ：To check the accuracy of our solution, we compute the cube of \(x_{3}\), which is approximately 4. This confirms that our implementation of Newton's method is correct.

Step 6 ：Final Answer: Therefore, the cube root of 4 is approximately \(\boxed{1.587401052148227}\).