Newton's Iteration Formula:
Newton's iteration formula is a method for finding the roots of a function. It is based on the idea of approximating the root using an initial guess and then refining the guess through a series of iterations. The formula for Newton's iteration is given by:

\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]

where \(x_n\) is the current approximation, \(f(x_n)\) is the value of the function at \(x_n\), and \(f'(x_n)\) is the derivative of the function at \(x_n\).

Applying Newton's Iteration Formula:
In this case, we need to find the cube root of 11. Let's use Newton's iteration formula to approximate the cube root of 11.

We can define our function as \(f(x) = x^3 - 11\), and the derivative of the function as \(f'(x) = 3x^2\).

Iteration 1:
Let's start with an initial guess of \(x_0 = 3\).

Using the iteration formula, we can calculate the next approximation:

\[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\]

Substituting the values:

\[x_1 = 3 - \frac{(3)^3 - 11}{3(3)^2}\]

Simplifying the equation:

\[x_1 = 3 - \frac{27 - 11}{27}\]

\[x_1 = 3 - \frac{16}{27}\]

\[x_1 = \frac{81}{27} - \frac{16}{27}\]

\[x_1 = \frac{65}{27}\]

Iteration 2:
Now, we can use the value of \(x_1\) to calculate the next approximation:

\[x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}\]

Substituting the values:

\[x_2 = \frac{65}{27} - \frac{(\frac{65}{27})^3 - 11}{3(\frac{65}{27})^2}\]

Simplifying the equation:

\[x_2 = \frac{65}{27} - \frac{\frac{274625}{19683} - 11}{3(\frac{4225}{729})}\]

\[x_2 = \frac{65}{27} - \frac{\frac{274625}{19683} - \frac{23653}{2187}}{\frac{4225}{243}}\]

\[x_2 = \frac{65}{27} - \frac{\frac{274625 - 23653(19683)}{19683}}{\frac{4225}{243}}\]

\[x_2 = \frac{65}{27} - \frac{\frac{274625 - 23653(19683)}{19683}}{\frac{4225}{243}}\]

\[x_2 = \frac{65}{27} - \frac{\frac{274625 - 464253269}{19683}}{\frac{4225}{243}}\]

\[x