Let $\alpha, \beta, \gamma$ be three numbers such that $\alpha + \beta + \gamma = 2$, $\alpha^2 + \beta^2 + \gamma^2 = 6$ and $\alpha^3 + \beta^3 + \gamma^3 = 11$, then:
Step-by-Step Solution
Key Concept: Newton's identities relate power sums of roots to elementary symmetric polynomials and allow recursive computation.
Using Vieta's formulas, $\sum \alpha\beta = \frac{2^2 - (-6)}{2} = -1$ and $11 - 3a\beta\gamma = 2(6-1) = 14$, giving $a\beta\gamma = -1$. From $x^2 - 2x^2 - x + 1 = 0$, we derive $x^3 = 2x^2 + x - 1$ for the roots. Computing higher powers and symmetric sums: $\alpha^4 + \beta^4 + \gamma^4 = 26$, $\alpha^5 + \beta^5 + \gamma^5 = 57$, and $\alpha^6 + \beta^6 + \gamma^6 = 129$. Finally, $-(4-\alpha^2)(4-\beta^2)(4-\gamma^2) = -2(2-\alpha)(2-\beta)(2-\gamma)(2+\alpha)(2+\beta)(2+\gamma) = 13$.
Correct Answer: [A-Q] [B-R] [C-P] [D-T]