1. **State the problem:** Solve the cubic equation $$x^3 - 6x^2 + 11x - 6 = 0$$. 2. **Recall the formula and approach:** For cubic equations, one common method is to try factoring by grouping or use the Rational Root Theorem to find possible roots. 3. **Apply the Rational Root Theorem:** Possible rational roots are factors of the constant term 6 divided by factors of the leading coefficient 1, i.e., $$\pm1, \pm2, \pm3, \pm6$$. 4. **Test possible roots:** - Test $$x=1$$: $$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$$, so $$x=1$$ is a root. 5. **Factor out $$x-1$$:** Use polynomial division or synthetic division to divide the cubic by $$x-1$$: $$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$$. 6. **Factor the quadratic:** $$x^2 - 5x + 6 = (x - 2)(x - 3)$$. 7. **Write the complete factorization:** $$x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$$. 8. **Find the roots:** Set each factor equal to zero: - $$x - 1 = 0 \Rightarrow x = 1$$ - $$x - 2 = 0 \Rightarrow x = 2$$ - $$x - 3 = 0 \Rightarrow x = 3$$ **Final answer:** The solutions to the equation are $$x = 1, 2, 3$$.