The value of α(β² + γ²) + β(γ² + α²) + γ(α² + β²) is divisible by -
Step-by-Step Solution
Key Concept: The expression \alpha(\beta^2 + \gamma^2) + \beta(\gamma^2 + \alpha^2) + \gamma(\alpha^2 + \beta^2) can be factored using the roots of the given cubic equation x^3 - 14x^2 + Px - 36 = 0. Given the determinant condition, the roots \alpha, \beta, \gamma satisfy \alpha + \beta + \gamma = 14, \alpha\beta + \beta\gamma + \gamma\alpha = P, and \alpha\beta\gamma = 36. The expression is equal to (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma = 14P - 3(36) = 14P - 108. Since t=2 is an even prime number, the determinant value is 2. Evaluating the determinant for the given matrix yields P=15. Thus, 14(15) - 108 = 210 - 108 = 102, which is divisible by 51.
The given determinant is equal to (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) = t = 2. The roots \alpha, \beta, \gamma satisfy x^3 - 14x^2 + Px - 36 = 0. From Vieta's formulas, \alpha+\beta+\gamma=14, \alpha\beta+\beta\gamma+\gamma\alpha=P, \alpha\beta\gamma=36. The expression E = \alpha(\beta^2+\gamma^2) + \beta(\gamma^2+\alpha^2) + \gamma(\alpha^2+\beta^2) = \alpha(\beta^2+\gamma^2+\alpha^2-\alpha^2) + \beta(\gamma^2+\alpha^2+\beta^2-\beta^2) + \gamma(\alpha^2+\beta^2+\gamma^2-\gamma^2) = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2) - (\alpha^3+\beta^3+\gamma^3). Using the identity \alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha)), we get E = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma = 14P - 3(36) = 14P - 108. Given the determinant value is 2, and solving for P, we find P=15. Thus E = 14(15) - 108 = 210 - 108 = 102. 102 is divisible by 51.
Correct Answer: C