Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 + ax^2 + bx + c = 0 \), then \( (\alpha + \beta - 2\gamma)(\beta + \gamma - 2\alpha)(\gamma + \alpha - 2\beta) = \)

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

Hint:

For symmetric expressions involving roots of a polynomial, using Vieta's formulas is key. When terms like \( \alpha+\beta-2\gamma \) appear, substitute \( \alpha+\beta = (\alpha+\beta+\gamma) - \gamma = -a - \gamma \) to express them in terms of the sum of all roots and a single root. This transforms the expression into a product of terms of the form \( (-a-3\alpha) \). This product can then be related to the polynomial whose roots are \( -3\alpha, -3\beta, -3\gamma \), which can be obtained by a suitable transformation of the original polynomial.

Answer 1

The Correct Option is D

Given the cubic equation \( x^3 + ax^2 + bx + c = 0 \), and \( \alpha, \beta, \gamma \) are its roots. From Vieta's formulas, we have: 1. Sum of the roots: \( \alpha + \beta + \gamma = -a \) 2. Sum of the products of roots taken two at a time: \( \alpha\beta + \beta\gamma + \gamma\alpha = b \) 3. Product of the roots: \( \alpha\beta\gamma = -c \) We need to evaluate the expression \( P = (\alpha + \beta - 2\gamma)(\beta + \gamma - 2\alpha)(\gamma + \alpha - 2\beta) \). From \( \alpha + \beta + \gamma = -a \), we can write: \( \alpha + \beta = -a - \gamma \) \( \beta + \gamma = -a - \alpha \) \( \gamma + \alpha = -a - \beta \) Substitute these into the expression \( P \): \( P = ((-a - \gamma) - 2\gamma)((-a - \alpha) - 2\alpha)((-a - \beta) - 2\beta) \) \( P = (-a - 3\gamma)(-a - 3\alpha)(-a - 3\beta) \) \( P = -(a + 3\gamma) \cdot -(a + 3\alpha) \cdot -(a + 3\beta) \) \( P = -(a + 3\alpha)(a + 3\beta)(a + 3\gamma) \) Let \( Q(x) = (x+3\alpha)(x+3\beta)(x+3\gamma) \). This is a polynomial whose roots are \( -3\alpha, -3\beta, -3\gamma \). We know that if \( \alpha, \beta, \gamma \) are the roots of \( f(x) = x^3 + ax^2 + bx + c = 0 \), then \( f(x) = (x-\alpha)(x-\beta)(x-\gamma) \). Consider a polynomial \( g(y) \) whose roots are \( -3\alpha, -3\beta, -3\gamma \). If \( y = -3x \), then \( x = -y/3 \). Substitute \( x = -y/3 \) into the original equation: \( \left(-\frac{y}{3}\right)^3 + a\left(-\frac{y}{3}\right)^2 + b\left(-\frac{y}{3}\right) + c = 0 \) \( -\frac{y^3}{27} + a\frac{y^2}{9} - b\frac{y}{3} + c = 0 \) Multiply by -27 to clear the denominators and make the leading coefficient positive: \( y^3 - 3ay^2 + 9by - 27c = 0 \) The roots of this equation are \( -3\alpha, -3\beta, -3\gamma \). So, the polynomial \( (y - (-3\alpha))(y - (-3\beta))(y - (-3\gamma)) = (y+3\alpha)(y+3\beta)(y+3\gamma) \) is equal to \( y^3 - 3ay^2 + 9by - 27c \). We need to evaluate \( -(a + 3\alpha)(a + 3\beta)(a + 3\gamma) \). This is \( -Q(-a) \) where \( Q(y) = (y+3\alpha)(y+3\beta)(y+3\gamma) = y^3 - 3ay^2 + 9by - 27c \). So, \( P = -[ (-a)^3 - 3a(-a)^2 + 9b(-a) - 27c ] \) \( P = -[ -a^3 - 3a(a^2) - 9ab - 27c ] \) \( P = -[ -a^3 - 3a^3 - 9ab - 27c ] \) \( P = -[ -4a^3 - 9ab - 27c ] \) \( P = 4a^3 + 9ab + 27c \) Let's recheck the substitution. The expression is \( -(a+3\alpha)(a+3\beta)(a+3\gamma) \). This is \( -P_0(a) \) where \( P_0(x) = (x+3\alpha)(x+3\beta)(x+3\gamma) \). We constructed the polynomial \( y^3 - 3ay^2 + 9by - 27c = (y - (-3\alpha))(y - (-3\beta))(y - (-3\gamma)) \). Let \( P(y) = y^3 - 3ay^2 + 9by - 27c \). Then \( P(y) = (y+3\alpha)(y+3\beta)(y+3\gamma) \). So we need to evaluate \( -P(a) \). \( P(a) = (a)^3 - 3a(a)^2 + 9b(a) - 27c \) \( P(a) = a^3 - 3a^3 + 9ab - 27c \) \( P(a) = -2a^3 + 9ab - 27c \) Therefore, \( (\alpha + \beta - 2\gamma)(\beta + \gamma - 2\alpha)(\gamma + \alpha - 2\beta) = -P(a) \) \( = -(-2a^3 + 9ab - 27c) \) \( = 2a^3 - 9ab + 27c \)