<p>Given \(x^2 - x + 1 = 0\) with roots \(\alpha\) and \(\beta\), the value of \(\alpha^{101} + \beta^{107}\) is:</p>
Step-by-Step Solution
Key Concept: The roots of x² - x + 1 = 0 are the complex cube roots of unity (specifically ω and ω²), so they satisfy ω³ = 1, allowing you to reduce large powers modulo 3.
<p><strong>Step 1:</strong> Find the roots of x² - x + 1 = 0</p><p>Using the quadratic formula: x = (1 ± √(1-4))/2 = (1 ± i√3)/2</p><p>These are ω and ω² where ω = e^(2πi/3), the primitive cube roots of unity.</p><p><strong>Step 2:</strong> Verify the cube root property</p><p>From x² - x + 1 = 0, we get x² + 1 = x, and multiplying by (x+1): x³ - 1 = 0, so x³ = 1</p><p>Thus α³ = 1 and β³ = 1</p><p><strong>Step 3:</strong> Reduce the exponents modulo 3</p><p>101 = 3(33) + 2, so α^101 = α²</p><p>107 = 3(35) + 2, so β^107 = β²</p><p><strong>Step 4:</strong> Calculate α² + β²</p><p>From Vieta's formulas: α + β = 1 and αβ = 1</p><p>α² + β² = (α + β)² - 2αβ = 1² - 2(1) = -1</p><p>∴ Answer: C</p>
Correct Answer: C