1. **State the problem:** Find all complex solutions to the equation $$(z+1)^3 = (\overline{z} - 2 + \sqrt{3}i)^3.$$\n\n2. **Rewrite the equation:** Since both sides are cubes and equal, we can take cube roots on both sides, noting cube roots of unity: $$z+1 = (\overline{z} - 2 + \sqrt{3}i) \cdot \omega$$ where $$\omega^3 = 1$$ and $$\omega \in \{1, e^{2\pi i/3}, e^{4\pi i/3}\}.$$\n\n3. **Express $z$ and $\overline{z}$:** Let $$z = x + yi,$$ so $$\overline{z} = x - yi.$$\n\n4. **Write equations for each root of unity:**\n- For $$\omega = 1$$ : $$z+1 = \overline{z} - 2 + \sqrt{3}i.$$\n- For $$\omega = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$ : $$z+1 = (\overline{z} - 2 + \sqrt{3}i)\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right).$$\n- For $$\omega = e^{4\pi i/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$$ : $$z+1 = (\overline{z} - 2 + \sqrt{3}i)\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right).$$\n\n5. **Solve for each case:**\n\n**Case 1: $$\omega=1$$**\n\n$$x + yi + 1 = x - yi - 2 + \sqrt{3}i$$\nSplitting real and imaginary parts:\n- Real: $$x + 1 = x - 2 \implies 1 = -2,$$ which is false. So no solution in this case.\n\n**Case 2: $$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$**\nLet $$w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad c = \overline{z} - 2 + \sqrt{3}i = (x - yi) - 2 + \sqrt{3}i = (x-2) + (-y + \sqrt{3})i.$$\nThen $$z+1 = cw.$$\nWrite $$cw = (x-2)(-\frac{1}{2}) - y(-\frac{\sqrt{3}}{2}) + i\left((x-2)\frac{\sqrt{3}}{2} + (-y + \sqrt{3})(-\frac{1}{2})\right).$$\nSimplify real and imaginary parts:\n- Real: $$-(x-2)/2 + y\sqrt{3}/2 = x + 1,$$\n- Imaginary: $$(x-2)\sqrt{3}/2 - (y - \sqrt{3})/2 = y.$$\nMultiply both equations by 2 for simplicity:\n\* Real: $$-x + 2 + y\sqrt{3} = 2x + 2,$$\n\* Imaginary: $$(x - 2)\sqrt{3} - y + \sqrt{3} = 2y.$$\nSimplify:\n- Real: $$-x + y\sqrt{3} + 2 = 2x + 2 \implies -x + y\sqrt{3} = 2x,$$\n- Imaginary: $$x\sqrt{3} - 2\sqrt{3} - y + \sqrt{3} = 2y \implies x\sqrt{3} - y - 2\sqrt{3} + \sqrt{3} = 2y \implies x\sqrt{3} - 2\sqrt{3} = 3y.$$\nFrom real: $$-x + y\sqrt{3} = 2x \implies y\sqrt{3} = 3x \implies y = \frac{3x}{\sqrt{3}} = x\sqrt{3}.$$\nFrom imaginary: $$x\sqrt{3} - 2\sqrt{3} = 3y.$$ Substitute $$y = x\sqrt{3}$$:\n$$x\sqrt{3} - 2\sqrt{3} = 3(x\sqrt{3}) = 3x\sqrt{3}.$$\nDivide by $$\sqrt{3}$$:\n$$x - 2 = 3x \implies -2 = 2x \implies x = -1.$$\nThen $$y = -1 \cdot \sqrt{3} = -\sqrt{3}.$$\n\nCheck $$z = -1 - \sqrt{3} i.$$\n\n**Case 3: $$\omega = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$$** is conjugate of Case 2, likewise:\n$$w = -\frac{1}{2} - \frac{\sqrt{3}}{2}i,$$ and $$cz = (x-2) + (-y + \sqrt{3})i.$$\nCalculate $$(c)(w)$$ real and imaginary parts similarly:\n- Real: $$-(x-2)/2 - y\sqrt{3}/2 = x + 1,$$\n- Imaginary: $$-(x-2)\sqrt{3}/2 - (y - \sqrt{3})/2 = y.$$\nMultiply by 2:\n- Real: $$-x + 2 - y\sqrt{3} = 2x + 2,$$\n- Imaginary: $$-(x-2)\sqrt{3} - y + \sqrt{3} = 2y.$$\nSimplify:\n- Real: $$-x - y\sqrt{3} + 2 = 2x + 2 \implies -x - y\sqrt{3} = 2x,$$\n- Imaginary: $$-x\sqrt{3} + 2\sqrt{3} - y + \sqrt{3} = 2y \implies -x\sqrt{3} + 3\sqrt{3} = 3y.$$\nFrom real:\n$$-x - y\sqrt{3} = 2x \implies -y\sqrt{3} = 3x \implies y = -\frac{3x}{\sqrt{3}} = -x\sqrt{3}.$$\nFrom imaginary:\n$$-x\sqrt{3} + 3\sqrt{3} = 3y = 3(-x\sqrt{3}) = -3x\sqrt{3}.$$\nDivide by $$\sqrt{3}$$:\n$$-x + 3 = -3x \implies 3 = -2x \implies x = -\frac{3}{2}.$$\nThen $$y = -x \sqrt{3} = -(-\frac{3}{2})\sqrt{3} = \frac{3\sqrt{3}}{2}.$$\n\nSo, $$z = -\frac{3}{2} + \frac{3\sqrt{3}}{2}i.$$\n\n**Final solutions:**\n\n$$\boxed{z = -1 - \sqrt{3} i \quad \text{and} \quad z = -\frac{3}{2} + \frac{3\sqrt{3}}{2} i}.$$\n\nNo solution from $$\omega = 1$$ case.