1. **State the problem:** Solve the cubic equation $$x^3 - 2x^2 + 3x - 6 = 0$$ and find all solutions. 2. **Recall the approach:** For cubic equations, try to find rational roots using the Rational Root Theorem, then factor and solve the remaining quadratic if possible. 3. **Check possible rational roots:** Factors of constant term 6 are $$\pm1, \pm2, \pm3, \pm6$$. 4. **Test $$x=1$$:** $$1^3 - 2(1)^2 + 3(1) - 6 = 1 - 2 + 3 - 6 = -4 \neq 0$$. 5. **Test $$x=2$$:** $$2^3 - 2(2)^2 + 3(2) - 6 = 8 - 8 + 6 - 6 = 0$$. So, $$x=2$$ is a root. 6. **Divide the cubic by $$x-2$$:** $$\frac{x^3 - 2x^2 + 3x - 6}{x-2} = x^2 + 3$$ 7. **Solve the quadratic $$x^2 + 3 = 0$$:** $$x^2 = -3$$ $$x = \pm \sqrt{-3} = \pm i\sqrt{3}$$ 8. **Final solutions:** $$x = 2, \quad x = i\sqrt{3}, \quad x = -i\sqrt{3}$$ These are one real root and two complex conjugate roots.