1. The problem is to solve the hardest question, but since no specific question was provided, I will demonstrate solving a challenging algebraic equation: Solve for $x$ in $$x^3 - 6x^2 + 11x - 6 = 0$$. 2. The formula used here is to find roots of a cubic polynomial. We can try to factor it by looking for rational roots using the Rational Root Theorem. 3. Possible rational roots are factors of the constant term 6: $\pm1, \pm2, \pm3, \pm6$. 4. Test $x=1$: $$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$$, so $x=1$ is a root. 5. Divide the polynomial by $(x-1)$ using synthetic division or polynomial division: $$x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6)$$ 6. Factor the quadratic $x^2 - 5x + 6$: $$x^2 - 5x + 6 = (x-2)(x-3)$$ 7. Therefore, the solutions are: $$x = 1, 2, 3$$ 8. These are the roots of the cubic equation, solving the problem completely.