1. **Problem 1:** Solve the equation $$f(x) = x^3 + x^2 - 3x - 3 = 0$$ using the False Position Method to find a positive real root. 2. **False Position Method Formula:** $$x_r = x_u - \frac{f(x_u)(x_l - x_u)}{f(x_l) - f(x_u)}$$ where $x_l$ and $x_u$ are the lower and upper guesses such that $f(x_l) \cdot f(x_u) < 0$. 3. **Step 1:** Choose initial guesses. Check values: - $f(1) = 1 + 1 - 3 - 3 = -4$ - $f(2) = 8 + 4 - 6 - 3 = 3$ Since $f(1) < 0$ and $f(2) > 0$, choose $x_l = 1$, $x_u = 2$. 4. **Step 2:** Calculate $x_r$: $$x_r = 2 - \frac{3(1 - 2)}{-4 - 3} = 2 - \frac{3(-1)}{-7} = 2 - \frac{-3}{-7} = 2 - \frac{3}{7} = 2 - 0.4286 = 1.5714$$ 5. **Step 3:** Evaluate $f(x_r)$: $$f(1.5714) = (1.5714)^3 + (1.5714)^2 - 3(1.5714) - 3 \approx 3.875 + 2.469 - 4.714 - 3 = 0.63$$ Since $f(x_r) > 0$ and $f(x_l) < 0$, update $x_u = x_r = 1.5714$. 6. **Step 4:** Repeat the process: Calculate new $x_r$: $$x_r = 1.5714 - \frac{0.63(1 - 1.5714)}{-4 - 0.63} = 1.5714 - \frac{0.63(-0.5714)}{-4.63} = 1.5714 - \frac{-0.36}{-4.63} = 1.5714 - 0.0777 = 1.4937$$ 7. **Step 5:** Evaluate $f(1.4937)$: $$f(1.4937) \approx 3.33 + 2.23 - 4.48 - 3 = 0.08$$ Since $f(x_r) > 0$, update $x_u = 1.4937$. 8. **Step 6:** Next iteration: $$x_r = 1.4937 - \frac{0.08(1 - 1.4937)}{-4 - 0.08} = 1.4937 - \frac{0.08(-0.4937)}{-4.08} = 1.4937 - \frac{-0.0395}{-4.08} = 1.4937 - 0.0097 = 1.4840$$ 9. **Step 7:** Evaluate $f(1.4840)$: $$f(1.4840) \approx 3.26 + 2.20 - 4.45 - 3 = 0.002$$ Close to zero, continue one more iteration. 10. **Step 8:** Next $x_r$: $$x_r = 1.4840 - \frac{0.002(1 - 1.4840)}{-4 - 0.002} = 1.4840 - \frac{0.002(-0.484)}{-4.002} = 1.4840 - \frac{-0.000968}{-4.002} = 1.4840 - 0.000242 = 1.4838$$ 11. **Step 9:** Evaluate $f(1.4838)$: $$f(1.4838) \approx 0.00001$$ Root is approximately $1.4838$. --- **Final answer for Problem 1:** The positive real root of $x^3 + x^2 - 3x - 3 = 0$ using False Position Method is approximately $$\boxed{1.4838}$$. --- (Note: Problem 2 is not solved as per instructions.)