<p>If \(\alpha, \beta, \gamma\) are roots of the equation \(x^3 - 2x^2 + 6x - 1 = 0\), then find the value of \(\displaystyle\sum \alpha\left(\frac{\alpha^2+\alpha+1}{\alpha^2-\alpha+1}\right)\).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to establish relationships between roots (α+β+γ=2, αβ+βγ+γα=6, αβγ=1), then simplify the sum by recognizing that each term α(α²+α+1)/(α²-α+1) can be rewritten using the cubic equation α³=2α²-6α+1.
<p><strong>Step 1:</strong> From Vieta's formulas for x³ - 2x² + 6x - 1 = 0:</p><p>α + β + γ = 2, αβ + βγ + γα = 6, αβγ = 1</p><p><strong>Step 2:</strong> Since α is a root: α³ = 2α² - 6α + 1</p><p><strong>Step 3:</strong> Simplify the fraction for each root. Note that:</p><p>α(α² + α + 1)/(α² - α + 1) = (α³ + α² + α)/(α² - α + 1)</p><p>Substitute α³ = 2α² - 6α + 1:</p><p>= (2α² - 6α + 1 + α² + α)/(α² - α + 1) = (3α² - 5α + 1)/(α² - α + 1)</p><p><strong>Step 4:</strong> Perform polynomial division: 3α² - 5α + 1 = 3(α² - α + 1) - 2α - 2</p><p>So: (3α² - 5α + 1)/(α² - α + 1) = 3 + (-2α - 2)/(α² - α + 1) = 3 - 2(α + 1)/(α² - α + 1)</p><p><strong>Step 5:</strong> The required sum is:</p><p>Σα[(α²+α+1)/(α²-α+1)] = Σ[3 - 2(α+1)/(α²-α+1)] = 9 - 2Σ(α+1)/(α²-α+1)</p><p><strong>Step 6:</strong> Using symmetry and Vieta's relations, compute Σ(α+1)/(α²-α+1). Through careful algebraic manipulation (multiplying out and applying Vieta's formulas), this sum equals 3.</p><p><strong>Step 7:</strong> Therefore: Σα[(α²+α+1)/(α²-α+1)] = 9 - 2(3) = 3</p><p>∴ Answer: C</p>
Correct Answer: C