Question

If the roots of the equation \(x^3 + ax^2 + bx + c = 0\) are in arithmetic progression, then:

• A. \(a^3 - 3ab + c = 0\)
• B. \(9ab = 2a^3 + 27c\)
• C. \(a^2 - 2bc + c = 0\)
• D. \(3ab - 3c - a^3 = 0\)

Hint:

When the roots of a polynomial are in arithmetic progression, leverage symmetry and basic algebraic identities like sum and product of roots to simplify the equations and find relationships between the coefficients.

Answer 1

The Correct Option is B

Step 1: Denote the roots by \( \alpha, \alpha+d, \alpha+2d \) due to the arithmetic progression.
The sum of the roots \( \alpha + (\alpha+d) + (\alpha+2d) = -a \), which simplifies to: \[ 3\alpha + 3d = -a \quad {or} \quad \alpha + d = -\frac{a}{3}. \] Step 2: The product of the roots is given by \( \alpha(\alpha+d)(\alpha+2d) = -c \).
Expanding this product: \[ \alpha(\alpha+d)(\alpha+2d) = \alpha^3 + 3\alpha^2 d + 2\alpha d^2 = -c. \] Step 3: Next, we express \( b \) in terms of \( \alpha \) and \( d \).
By Vieta's formulas, the sum of the products of the roots taken two at a time is \( b \): \[ \alpha(\alpha+d) + \alpha(\alpha+2d) + (\alpha+d)(\alpha+2d) = b. \] Simplify the terms: \[ \alpha(\alpha+d) = \alpha^2 + \alpha d, \quad \alpha(\alpha+2d) = \alpha^2 + 2\alpha d, \quad (\alpha+d)(\alpha+2d) = \alpha^2 + 3\alpha d + 2d^2. \] Adding these up: \[ 3\alpha^2 + 6\alpha d + 2d^2 = b. \] Step 4: Solve the relationships between \(a\), \(b\), and \(c\).
We substitute the expressions for the sum and product of the roots into the identity for the cubic polynomial. After some algebraic manipulation, we find that the correct relationship is: \[ 9ab = 2a^3 + 27c. \] Thus, the correct answer is: \[ \boxed{(B)} 9ab = 2a^3 + 27c. \]