1. **State the problem:** Solve the equation $$(x - 1)^3 + \frac{3}{2}(2x + 3)^2 + (-1 - x)^3 = 18x.$$\n\n2. **Rewrite the equation:** Note that $(-1 - x)^3 = -(x + 1)^3$. So the equation becomes:\n$$ (x - 1)^3 + \frac{3}{2}(2x + 3)^2 - (x + 1)^3 = 18x. $$\n\n3. **Expand the cubes:**\n$$ (x - 1)^3 = x^3 - 3x^2 + 3x - 1, $$\n$$ (x + 1)^3 = x^3 + 3x^2 + 3x + 1. $$\n\n4. **Expand the square:**\n$$ (2x + 3)^2 = 4x^2 + 12x + 9. $$\n\n5. **Substitute expansions back:**\n$$ (x^3 - 3x^2 + 3x - 1) + \frac{3}{2}(4x^2 + 12x + 9) - (x^3 + 3x^2 + 3x + 1) = 18x. $$\n\n6. **Distribute the $\frac{3}{2}$:**\n$$ (x^3 - 3x^2 + 3x - 1) + (6x^2 + 18x + \frac{27}{2}) - (x^3 + 3x^2 + 3x + 1) = 18x. $$\n\n7. **Combine like terms on the left:**\n$$ x^3 - 3x^2 + 3x - 1 + 6x^2 + 18x + \frac{27}{2} - x^3 - 3x^2 - 3x - 1 = 18x. $$\n\n8. **Simplify:**\n- $x^3 - x^3 = 0$\n- $-3x^2 + 6x^2 - 3x^2 = 0$\n- $3x + 18x - 3x = 18x$\n- $-1 + \frac{27}{2} - 1 = \frac{27}{2} - 2 = \frac{23}{2}$\n\nSo the left side simplifies to:\n$$ 18x + \frac{23}{2} = 18x. $$\n\n9. **Subtract $18x$ from both sides:**\n$$ 18x + \frac{23}{2} - 18x = 18x - 18x $$\n$$ \cancel{18x} + \frac{23}{2} - \cancel{18x} = 0 $$\n$$ \frac{23}{2} = 0. $$\n\n10. **Conclusion:** The equation reduces to $\frac{23}{2} = 0$, which is false.\n\nTherefore, there is **no solution** to the equation.