If the roots of the equation, $x^{3}+3a\mathrm{x}^{2}+3bx+c=0$ are in H.P. then
- $b^{2}=c(3ab-c)$
- $2b^{3}=c(3ab-c)$
- $2b^{3}=c^{2} (3 {\it ab-c})$
- $2b^{2}=c^{2} (3 {\it ab-c})$
If roots of x^3 + 3ax^2 + 3bx + c = 0 are in HP, their reciprocals are in AP. Let roots be 1/p, 1/q, 1/r. The equation for reciprocals is cx^3 + 3bx^2 + 3ax + 1 = 0. For roots in AP, 2 * (second term coefficient / first term coefficient) = ... leads to the condition 2b^3 = c(3ab - c).
Let the roots of the cubic equation be 1/(a-d), 1/a, and 1/(a+d). Since they are in harmonic progression, taking reciprocals gives roots in arithmetic progression for the transformed equation. The sum of the roots is -3a, so (1/(a-d) + 1/a + 1/(a+d)) equals -3a; however, it is easier to substitute y = 1/x into the original equation to get c(y^3) + 3b(y^2) + 3ay + 1 = 0. For this new cubic equation, the roots are in arithmetic progression, meaning the second root is the negative of the coefficient of y^2 divided by 3 times the leading coefficient, giving -b/c. Substituting y = -b/c into the cubic equation yields c(-b^3/c^3) + 3b(b^2/c^2) + 3a(-b/c) + 1 = 0. Multiplying by c^2 gives -b^3 + 3b^3 - 3abc + c^2 = 0, which simplifies to 2b^3 = c(3ab - c).