1. **State the problem:** We want to find an approximate solution to the equation $$x^3 - 4x + 1 = 0$$ using the iterative formula $$x_{n+1} = - \frac{1}{x_n^2 - 4}$$ starting with $$x_1 = 0.6$$. 2. **Explain the iterative method:** This method generates successive approximations $$x_2, x_3, \ldots$$ by plugging the previous value into the formula. 3. **Calculate $$x_2$$:** $$x_2 = - \frac{1}{(0.6)^2 - 4} = - \frac{1}{0.36 - 4} = - \frac{1}{-3.64} = 0.2747$$ (rounded to 4 d.p.) 4. **Calculate $$x_3$$:** $$x_3 = - \frac{1}{(0.2747)^2 - 4} = - \frac{1}{0.0755 - 4} = - \frac{1}{-3.9245} = 0.2549$$ 5. **Calculate $$x_4$$:** $$x_4 = - \frac{1}{(0.2549)^2 - 4} = - \frac{1}{0.0650 - 4} = - \frac{1}{-3.9350} = 0.2544$$ 6. **Calculate $$x_5$$:** $$x_5 = - \frac{1}{(0.2544)^2 - 4} = - \frac{1}{0.0647 - 4} = - \frac{1}{-3.9353} = 0.2544$$ 7. **Conclusion:** The values are converging to approximately $$0.254$$ to 3 decimal places. **Final answer:** $$x \approx 0.254$$