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The problem involves using an iteration formula to calculate values for \( x_1 \), \( x_2 \), and \( x_3 \), starting with \( x_0 = 2 \). The iteration formula is given by:

\[
x_{n+1} = \sqrt[3]{10 - 2x_n}
\]

Let's solve for \( x_1 \), \( x_2 \), and \( x_3 \) step by step.

### Step 1: Find \( x_1 \)
Starting with \( x_0 = 2 \):
\[
x_1 = \sqrt[3]{10 - 2(2)} = \sqrt[3]{10 - 4} = \sqrt[3]{6} \approx 1.817
\]

### Step 2: Find \( x_2 \)
Using \( x_1 \approx 1.817 \):
\[
x_2 = \sqrt[3]{10 - 2(1.817)} = \sqrt[3]{10 - 3.634} = \sqrt[3]{6.366} \approx 1.854
\]

### Step 3: Find \( x_3 \)
Using \( x_2 \approx 1.854 \):
\[
x_3 = \sqrt[3]{10 - 2(1.854)} = \sqrt[3]{10 - 3.708} = \sqrt[3]{6.292} \approx 1.849
\]

### Final values:
- \( x_1 \approx 1.817 \)
- \( x_2 \approx 1.854 \)
- \( x_3 \approx 1.849 \)

Would you like more details or explanations on any part of the solution?

Here are some related questions:
1. What happens if you change the initial value \( x_0 \)?
2. What are some uses of iterative methods in mathematics?
3. How can you determine when to stop iterations in this formula?
4. Can this method find solutions to any type of equation?
5. What other methods can be used to approximate solutions numerically?

**Tip**: When using iteration methods, always check if the values are converging towards a single value after several iterations.

Use the iteration formula x_(n+1) = ∛(10 - 2x_n) to find the values of x_1, x_2, and x_3. Start with x_0 = 2.

Solve an iteration formula problem starting with x_0 = 2. Learn how to calculate values for x_1, x_2, and x_3 using algebraic cube roots. This step-by-step solution is perfect for students in Grades 10-12.

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