1. The problem is to find an approximate solution to the equation $$x^3 + 7x = 1$$ using the iterative formula $$x_{n+1} = \frac{1 - x_n^3}{7}$$ starting with $$x_1 = 0$$ and to give the answer to 3 decimal places. 2. The iterative method uses the formula repeatedly to get closer to the solution. We start with $$x_1 = 0$$ and calculate subsequent values: 3. Calculate $$x_2$$: $$x_2 = \frac{1 - (0)^3}{7} = \frac{1}{7} = 0.142857$$ 4. Calculate $$x_3$$: $$x_3 = \frac{1 - (0.142857)^3}{7} = \frac{1 - 0.002915}{7} = \frac{0.997085}{7} = 0.142441$$ 5. Calculate $$x_4$$: $$x_4 = \frac{1 - (0.142441)^3}{7} = \frac{1 - 0.002889}{7} = \frac{0.997111}{7} = 0.142444$$ 6. Calculate $$x_5$$: $$x_5 = \frac{1 - (0.142444)^3}{7} = \frac{1 - 0.002889}{7} = \frac{0.997111}{7} = 0.142444$$ 7. The values have stabilized to about $$0.142$$ when rounded to 3 decimal places. 8. Therefore, the approximate solution to 3 decimal places is $$\boxed{0.142}$$.