1. **State the problem:** Solve the inequality $$t - 1(t - 6)(t + 1) < 0$$. 2. **Rewrite the inequality:** The expression is ambiguous, so we interpret it as $$t - 1 \cdot (t - 6)(t + 1) < 0$$ which simplifies to $$t - (t - 6)(t + 1) < 0$$. 3. **Expand the product:** $$(t - 6)(t + 1) = t^2 + t - 6t - 6 = t^2 - 5t - 6$$ 4. **Substitute back:** $$t - (t^2 - 5t - 6) < 0$$ 5. **Distribute the minus sign:** $$t - t^2 + 5t + 6 < 0$$ 6. **Combine like terms:** $$-t^2 + 6t + 6 < 0$$ 7. **Multiply both sides by -1 to make the quadratic coefficient positive, remembering to reverse the inequality:** $$\cancel{-1} \cdot (-t^2 + 6t + 6) > \cancel{-1} \cdot 0$$ $$t^2 - 6t - 6 > 0$$ 8. **Find the roots of the quadratic equation $$t^2 - 6t - 6 = 0$$ using the quadratic formula:** $$t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 + 24}}{2} = \frac{6 \pm \sqrt{60}}{2}$$ 9. **Simplify the square root:** $$\sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15}$$ 10. **Calculate the roots:** $$t = \frac{6 \pm 2\sqrt{15}}{2} = 3 \pm \sqrt{15}$$ 11. **Determine the intervals where $$t^2 - 6t - 6 > 0$$:** Since the parabola opens upward, the inequality holds outside the roots: $$t < 3 - \sqrt{15} \quad \text{or} \quad t > 3 + \sqrt{15}$$ 12. **Final answer:** $$\boxed{t < 3 - \sqrt{15} \quad \text{or} \quad t > 3 + \sqrt{15}}$$