<p>Let \(Q(x) = (x^3 - 2x^2 + x - 1)(x+1) = 0 \Rightarrow \delta = -1\).<br>\(\alpha + \beta + \gamma - 2,\ \alpha\beta + \beta\gamma + \gamma\alpha - 1\).<br>\(P(x) = (x^2+1)Q(x) + x^2 - x + 1\).<br>Find \(P(\alpha) + P(\beta) + P(\gamma) + P(-1)\).</p>
Step-by-Step Solution
Key Concept: Since α, β, γ are roots of Q(x) = 0, we have Q(α) = Q(β) = Q(γ) = 0, making P(α) = P(β) = P(γ) = α² - α + 1. The sum reduces to evaluating three identical terms plus P(-1).
<p><strong>Step 1:</strong> Identify roots. Q(x) = (x³ - 2x² + x - 1)(x+1) has roots α, β, γ from the cubic factor, and -1 from (x+1).</p><p><strong>Step 2:</strong> For roots of Q(x): Since Q(α) = 0, we have (α³ - 2α² + α - 1)(α+1) = 0. This means (x+1)P(x) = (x²+1)·0 + x² - x + 1, so P(α) = α² - α + 1. Same for β and γ.</p><p><strong>Step 3:</strong> From given conditions: α + β + γ = 2 and αβ + βγ + γα = 1 (computed from Vieta's formulas on x³ - 2x² + x - 1 = 0).</p><p><strong>Step 4:</strong> Calculate sum for roots of cubic:<br>P(α) + P(β) + P(γ) = (α² - α + 1) + (β² - β + 1) + (γ² - γ + 1)<br>= (α² + β² + γ²) - (α + β + γ) + 3<br>= [(α+β+γ)² - 2(αβ+βγ+γα)] - (α+β+γ) + 3<br>= [4 - 2(1)] - 2 + 3 = 2 - 2 + 3 = 3</p><p><strong>Step 5:</strong> Calculate P(-1):<br>P(-1) = ((-1)²+1)Q(-1) + (-1)² - (-1) + 1<br>= 2·0 + 1 + 1 + 1 = 3</p><p><strong>Step 6:</strong> Final sum = 3 + 3 = <strong>6</strong></p>
Correct Answer: 6