1. The problem is to find an approximate solution to the equation $$x^3 - 3x + 1 = 0$$ using the iterative formula $$x_{n+1} = - \frac{1}{x_n^2 - 3}$$ starting with $$x_1 = 0.5$$. 2. The iterative method involves substituting the current value $$x_n$$ into the formula to find the next value $$x_{n+1}$$. 3. Calculate $$x_2$$: $$x_2 = - \frac{1}{(0.5)^2 - 3} = - \frac{1}{0.25 - 3} = - \frac{1}{-2.75} = 0.3636$$ (rounded to 4 d.p.) 4. Calculate $$x_3$$: $$x_3 = - \frac{1}{(0.3636)^2 - 3} = - \frac{1}{0.1322 - 3} = - \frac{1}{-2.8678} = 0.3488$$ (rounded to 4 d.p.) 5. Calculate $$x_4$$: $$x_4 = - \frac{1}{(0.3488)^2 - 3} = - \frac{1}{0.1217 - 3} = - \frac{1}{-2.8783} = 0.3475$$ (rounded to 4 d.p.) 6. Calculate $$x_5$$: $$x_5 = - \frac{1}{(0.3475)^2 - 3} = - \frac{1}{0.1207 - 3} = - \frac{1}{-2.8793} = 0.3474$$ (rounded to 4 d.p.) 7. The values are converging, so the approximate solution to 3 decimal places is $$x \approx 0.347$$. Final answer: $$\boxed{0.347}$$