Question

If the sum of the cubes of the roots of the equation \( x^3 - ax^2 + bx - c = 0 \) is zero, then \( a^3 + 3c = \):

• A. \( -2ab \)
• B. \( 2ab \)
• C. \( -3ab \)
• D. \( 3ab \)

Hint:

For cubic equations, use Vieta's formulas to express the sum and product of roots, and apply relevant identities to solve for unknowns.

Answer 1

The Correct Option is D

Let the roots of the equation \( x^3 - ax^2 + bx - c = 0 \) be \( \alpha, \beta, \gamma \). According to Vieta's formulas: \[ \alpha + \beta + \gamma = a, \quad \alpha\beta + \beta\gamma + \gamma\alpha = b, \quad \alpha\beta\gamma = c \] We are given that the sum of the cubes of the roots is zero: \[ \alpha^3 + \beta^3 + \gamma^3 = 0 \] Using the identity: \[ \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma) \left( (\alpha + \beta + \gamma)^2 - 3(\alpha\beta + \beta\gamma + \gamma\alpha) \right) - 3\alpha\beta\gamma \] Substitute the known values from Vieta's formulas: \[ 0 = a \left( a^2 - 3b \right) - 3c \] Thus, we have: \[ a(a^2 - 3b) = 3c \] This simplifies to: \[ a^3 - 3ab = 3c \] Therefore, \( a^3 + 3c = 3ab \).