<p>If <em>α</em>, <em>β</em> ∈ <em>C</em> are the distinct roots of the equation \(x^2 - x + 1 = 0\), then \(\alpha^{101} + \beta^{107}\) is equal to</p>
Step-by-Step Solution
Key Concept: The roots of x² - x + 1 = 0 are the primitive 6th roots of unity (specifically ω and ω²). Use the periodicity property: these roots satisfy α³ = -1 and β³ = -1, or equivalently α⁶ = 1, so reduce exponents modulo 6.
<p><strong>Step 1: Find the roots</strong></p><p>For x² - x + 1 = 0: x = (1 ± √(1-4))/2 = (1 ± i√3)/2</p><p>These are ω = e^(iπ/3) and ω² = e^(i2π/3), where ω⁶ = 1.</p><p><strong>Step 2: Verify α and β are 6th roots of unity</strong></p><p>From x² - x + 1 = 0, we get x² = x - 1. Multiply by (x+1):</p><p>(x+1)(x² - x + 1) = x(x-1) + (x-1) = x³ + 1 = 0</p><p>So x³ = -1, which means x⁶ = 1. Also: α³ = -1 and β³ = -1</p><p><strong>Step 3: Reduce exponents modulo 6</strong></p><p>101 = 16(6) + 5, so α¹⁰¹ = α⁵</p><p>107 = 17(6) + 5, so β¹⁰⁷ = β⁵</p><p><strong>Step 4: Calculate α⁵ + β⁵</strong></p><p>Since α³ = -1: α⁵ = α³ · α² = -α²</p><p>Since β³ = -1: β⁵ = β³ · β² = -β²</p><p>Therefore: α¹⁰¹ + β¹⁰⁷ = -α² - β² = -(α² + β²)</p><p><strong>Step 5: Use Vieta's formulas</strong></p><p>α + β = 1 and αβ = 1</p><p>α² + β² = (α + β)² - 2αβ = 1 - 2 = -1</p><p>∴ α¹⁰¹ + β¹⁰⁷ = -(-1) = <strong>1</strong></p>
Correct Answer: D