If the equation $x^3 - 3ax^2 + 3bx- c = 0$ has positive and distinct roots, then
- $a^2 > b$
- $ab > c$
- $a^3 > c$
- $a^3 > b^2 > c$
For cubic x^3 - 3ax^2 + 3bx - c = 0, if roots are p, q, r, then p+q+r = 3a, pq+qr+rp = 3b, pqr = c. For distinct real roots, the AM-GM inequality or discriminant properties apply. Specifically, a^2 > b is a necessary condition for real roots in this form.
If the distinct and positive roots are p, q, and r, then by Vieta's formulas for the cubic x^3 - 3ax^2 + 3bx - c = 0, the sum of the roots p+q+r equals 3a, and the sum of their pairwise products pq+qr+rp equals 3b. Using the algebraic inequality, the square of the sum of positive variables must strictly exceed their pairwise product sum, giving (p+q+r)^2 > 3(pq+qr+rp). Substituting the Vieta's formulas into this inequality yields (3a)^2 > 3(3b), which simplifies directly to a^2 > b.