1. **State the problem:** Find all complex numbers $z$ such that $$ (z+1)^3 = (\overline{z} - 2 + \sqrt{3}i)^3. $$ 2. **Apply the cube root:** Since both sides are cubes, we equate the bases up to multiplication by a cube root of unity. Let $\omega$ be a cube root of unity, satisfying $\omega^3 = 1$. Then the solutions satisfy $$ z + 1 = \omega^k (\overline{z} - 2 + \sqrt{3} i) \quad \text{for } k=0,1,2. $$ 3. **Case 1: $k=0$** $$ z + 1 = \overline{z} - 2 + \sqrt{3} i. $$ Let $z = x + yi$, then $\overline{z} = x - yi$. Substitute: $$ (x + yi) + 1 = (x - yi) - 2 + \sqrt{3} i. $$ Equate real and imaginary parts: - Real: $x + 1 = x - 2$ which implies $1 = -2$, contradiction. So no solutions in case 1. 4. **Case 2: $k=1$** $$ z + 1 = \omega (\overline{z} - 2 + \sqrt{3} i). $$ Recall $\omega = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$. Write explicitly: $$ z + 1 = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)(\overline{z} - 2 + \sqrt{3} i). $$ Let $z = x + yi$, $\overline{z} = x - yi$. The right side is $$ \left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)(x - yi - 2 + \sqrt{3} i). $$ Simplify inside parentheses: $$ (x - 2) + (-y + \sqrt{3}) i. $$ Multiply: Real part = $$ -\frac{1}{2} (x - 2) - \frac{\sqrt{3}}{2} (-y + \sqrt{3}) = -\frac{x}{2} + 1 + \frac{\sqrt{3} y}{2} - \frac{3}{2}, $$ Imag part = $$ \frac{\sqrt{3}}{2} (x - 2) - \frac{1}{2}(-y + \sqrt{3}) = \frac{\sqrt{3} x}{2} - \sqrt{3} + \frac{y}{2} - \frac{\sqrt{3}}{2}. $$ Simplify real: $$ -\frac{x}{2} + 1 + \frac{\sqrt{3} y}{2} - \frac{3}{2} = -\frac{x}{2} + \frac{\sqrt{3} y}{2} - \frac{1}{2}. $$ Simplify imaginary: $$ \frac{\sqrt{3} x}{2} + \frac{y}{2} - \frac{3\sqrt{3}}{2}. $$ Set left side $z + 1 = (x + 1) + yi$ equal to this: Equate real parts: $$ x + 1 = -\frac{x}{2} + \frac{\sqrt{3} y}{2} - \frac{1}{2}, $$ Equate imaginary parts: $$ y = \frac{\sqrt{3} x}{2} + \frac{y}{2} - \frac{3 \sqrt{3}}{2}. $$ 5. **Solve the system:** From real parts: $$ x + 1 + \frac{x}{2} = \frac{\sqrt{3} y}{2} - \frac{1}{2} \implies \frac{3x}{2} + \frac{3}{2} = \frac{\sqrt{3} y}{2} \implies 3x + 3 = \sqrt{3} y. $$ From imaginary parts: $$ y - \frac{y}{2} = \frac{\sqrt{3} x}{2} - \frac{3 \sqrt{3}}{2} \implies \frac{y}{2} = \frac{\sqrt{3} x}{2} - \frac{3 \sqrt{3}}{2} \implies y = \sqrt{3} x - 3 \sqrt{3}. $$ Substitute $y$ into the first equation: $$ 3x + 3 = \sqrt{3} (\sqrt{3} x - 3 \sqrt{3}) = 3x - 3 \times 3 = 3x - 9. $$ Then $$ 3x + 3 = 3x - 9 \implies 3 = -9, $$ contradiction. No solutions for $k=1$. 6. **Case 3: $k=2$** $$ z + 1 = \omega^2 (\overline{z} - 2 + \sqrt{3} i) $$ where $\omega^2 = e^{4\pi i /3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i$. Proceed as in case 2: $$ z + 1 = \left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)(x - 2 + (-y + \sqrt{3}) i). $$ Calculate real part: $$ -\frac{1}{2} (x - 2) + \frac{\sqrt{3}}{2} (-y + \sqrt{3}) = -\frac{x}{2} + 1 - \frac{\sqrt{3} y}{2} + \frac{3}{2} = -\frac{x}{2} - \frac{\sqrt{3} y}{2} + \frac{5}{2}. $$ Imaginary part: $$ -\frac{\sqrt{3}}{2} (x - 2) - \frac{1}{2} (-y + \sqrt{3}) = -\frac{\sqrt{3} x}{2} + \sqrt{3} + \frac{y}{2} - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3} x}{2} + \frac{y}{2} + \frac{\sqrt{3}}{2}. $$ Set equal to $z + 1 = (x + 1) + yi$: Real: $$ x + 1 = -\frac{x}{2} - \frac{\sqrt{3} y}{2} + \frac{5}{2}, $$ Imag: $$ y = -\frac{\sqrt{3} x}{2} + \frac{y}{2} + \frac{\sqrt{3}}{2}. $$ 7. **Solve system for case 3:** From real parts: $$ x + 1 + \frac{x}{2} = -\frac{\sqrt{3} y}{2} + \frac{5}{2} \implies \frac{3x}{2} + 1 = -\frac{\sqrt{3} y}{2} + \frac{5}{2} \implies 3x - \sqrt{3} y = 4. $$ From imaginary parts: $$ y - \frac{y}{2} = -\frac{\sqrt{3} x}{2} + \frac{\sqrt{3}}{2} \implies \frac{y}{2} = -\frac{\sqrt{3} x}{2} + \frac{\sqrt{3}}{2} \implies y = -\sqrt{3} x + \sqrt{3}. $$ Substitute $y$ into the first equation: $$ 3x - \sqrt{3} (-\sqrt{3} x + \sqrt{3}) = 4 \implies 3x + 3x - 3 = 4 \implies 6x = 7 \implies x = \frac{7}{6}. $$ Then $$ y = -\sqrt{3} \times \frac{7}{6} + \sqrt{3} = \sqrt{3} \left(1 - \frac{7}{6}\right) = \sqrt{3} \times \left(-\frac{1}{6}\right) = -\frac{\sqrt{3}}{6}. $$ 8. **Final answer:** $$ z = x + yi = \frac{7}{6} - \frac{\sqrt{3}}{6} i. $$ No other solutions exist. **Summary:** The only complex solution to the given equation is $$ z = \frac{7}{6} - \frac{\sqrt{3}}{6} i. $$