To solve the given equation $x^3 - x - 1 = 0$ using the Newton-Raphson method starting with $x_0 = 1.5$, and performing three iterations, here are the steps:

Step-by-Step Process:

1. Newton-Raphson Formula:

The Newton-Raphson iteration formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

2. Define the Function and Its Derivative:

The given function is: $f(x) = x^3 - x - 1$ Its derivative is: $f'(x) = 3x^2 - 1$

3. Initial Guess:

Starting with $x_0 = 1.5$.

Iteration 1:

$f(x_0) = f(1.5) = (1.5)^3 - 1.5 - 1 = 3.375 - 1.5 - 1 = 0.875$ $f'(x_0) = f'(1.5) = 3(1.5)^2 - 1 = 3(2.25) - 1 = 6.75 - 1 = 5.75$ Using the formula: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.5 - \frac{0.875}{5.75} = 1.5 - 0.15217 = 1.34783$

Iteration 2:

Using $x_1 = 1.34783$: $f(x_1) = f(1.34783) = (1.34783)^3 - 1.34783 - 1 = 2.4485 - 1.34783 - 1 = 0.10067$ $f'(x_1) = f'(1.34783) = 3(1.34783)^2 - 1 = 3(1.81689) - 1 = 5.45067 - 1 = 4.45067$ Using the formula: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.34783 - \frac{0.10067}{4.45067} = 1.34783 - 0.02262 = 1.32521$

Iteration 3:

Using $x_2 = 1.32521$: $f(x_2) = f(1.32521) = (1.32521)^3 - 1.32521 - 1 = 2.32889 - 1.32521 - 1 = 0.00368$ $f'(x_2) = f'(1.32521) = 3(1.32521)^2 - 1 = 3(1.75523) - 1 = 5.26569 - 1 = 4.26569$ Using the formula: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.32521 - \frac{0.00368}{4.26569} = 1.32521 - 0.00086 = 1.32435$

Final Approximation:

After three iterations, the approximate root is: $x_3 \approx 1.32435$

Would you like a detailed explanation of any specific step? Here are some related questions to explore further:

1. Why is the Newton-Raphson method efficient for finding roots?
2. What happens if the initial guess is far from the actual root?
3. How do you know the solution is accurate enough after three iterations?
4. Can this method fail under certain conditions? If so, why?
5. How does the Newton-Raphson method compare to other root-finding methods?

Tip: Always check $f'(x)$ before starting Newton-Raphson to avoid division by zero or near-zero values.