1. **State the problem:** Solve the inequality $$x^3 - 6x^2 + 11x - 6 < 0$$. 2. **Factor the cubic polynomial:** To solve the inequality, first factor the cubic expression. We look for rational roots using the Rational Root Theorem. Possible roots are factors of 6: $\pm1, \pm2, \pm3, \pm6$. 3. **Test roots:** - For $x=1$: $$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$$ so $x=1$ is a root. 4. **Divide the polynomial by $(x-1)$:** $$\frac{x^3 - 6x^2 + 11x - 6}{x-1} = x^2 - 5x + 6$$ 5. **Factor the quadratic:** $$x^2 - 5x + 6 = (x-2)(x-3)$$ 6. **Complete factorization:** $$x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$$ 7. **Analyze the inequality:** We want to find where $$(x-1)(x-2)(x-3) < 0$$ 8. **Determine sign intervals:** - For $x < 1$, all three factors are negative or positive? - $(x-1)<0$, $(x-2)<0$, $(x-3)<0$ so product is $(-)(-)(-) = -$ negative. - For $1 < x < 2$: - $(x-1)>0$, $(x-2)<0$, $(x-3)<0$ so product is $(+)(-)(-) = +$ positive. - For $2 < x < 3$: - $(x-1)>0$, $(x-2)>0$, $(x-3)<0$ so product is $(+)(+)(-) = -$ negative. - For $x > 3$: - $(x-1)>0$, $(x-2)>0$, $(x-3)>0$ so product is $(+)(+)(+) = +$ positive. 9. **Solution:** The inequality holds where the product is negative: $$x < 1 \quad \text{or} \quad 2 < x < 3$$ **Final answer:** $$(-\infty, 1) \cup (2, 3)$$