1. **State the problem:** We are given the function $f(x) = x^3 - 3x + 1$ and initial points $x_0 = 0$ and $x_1 = 1$. We need to use the method of False Position (Regula Falsi) to find $x_2$ and $x_3$. 2. **Recall the method of False Position formula:** $$x_{n+1} = x_n - f(x_n) \cdot \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$$ This method uses a secant line between points $(x_{n-1}, f(x_{n-1}))$ and $(x_n, f(x_n))$ to approximate the root. 3. **Calculate $f(x_0)$ and $f(x_1)$:** $$f(0) = 0^3 - 3\cdot0 + 1 = 1$$ $$f(1) = 1^3 - 3\cdot1 + 1 = 1 - 3 + 1 = -1$$ 4. **Calculate $x_2$ using the formula:** $$x_2 = 1 - (-1) \cdot \frac{1 - 0}{-1 - 1} = 1 - (-1) \cdot \frac{1}{-2} = 1 - \frac{-1}{-2} = 1 - \frac{1}{2} = 0.5$$ 5. **Calculate $f(x_2)$:** $$f(0.5) = (0.5)^3 - 3\cdot0.5 + 1 = 0.125 - 1.5 + 1 = -0.375$$ 6. **Determine the new interval:** Since $f(0) = 1$ and $f(0.5) = -0.375$, the root lies between $0$ and $0.5$ because the function values have opposite signs. 7. **Calculate $x_3$ using points $x_0=0$ and $x_2=0.5$:** $$x_3 = 0.5 - (-0.375) \cdot \frac{0.5 - 0}{-0.375 - 1} = 0.5 - (-0.375) \cdot \frac{0.5}{-1.375} = 0.5 - \frac{-0.1875}{-1.375} = 0.5 - 0.13636 = 0.36364$$ 8. **Summary:** - $x_2 = 0.5$ - $x_3 \approx 0.36364$ These are the next two approximations of the root using the method of False Position.