<p>If \(x^2 - x + 1 = 0\) has roots \(\alpha\) and \(\beta\), then the value of \(\alpha^{2009} + \beta^{2009}\) 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), which satisfy ω³ = 1. Use this periodicity to reduce the exponent 2009 modulo 3.
<p><strong>Step 1: Identify the roots</strong><br/>From x² - x + 1 = 0, using the quadratic formula:</p><p>α, β = (1 ± √(1-4))/2 = (1 ± i√3)/2</p><p>These are ω and ω² where ω = e^(2πi/3) (primitive cube root of unity).</p><p><strong>Step 2: Verify ω³ = 1</strong><br/>Since x² - x + 1 = 0, we have x² + 1 = x, so x³ - 1 = (x-1)(x² + x + 1) = 0.<br/>Actually, multiply x² - x + 1 = 0 by (x+1): x³ + 1 = 0, giving x³ = -1.<br/>Correction: From x² - x + 1 = 0, multiply by (x+1): (x+1)(x²-x+1) = x³+1, so α³ = -1.<br/>Better: Note that 1 + ω + ω² = 0 and ω³ = 1 are properties of primitive cube roots.</p><p><strong>Step 3: Use periodicity</strong><br/>Since α³ = -1 (or equivalently α⁶ = 1), we have period 6.<br/>2009 = 6(334) + 5<br/>Therefore: α^2009 = α^5 and β^2009 = β^5</p><p><strong>Step 4: Calculate α^5 + β^5</strong><br/>From x² - x + 1 = 0: α + β = 1 and αβ = 1<br/>α² = α - 1 (from the equation)<br/>α³ = α·α² = α(α-1) = α² - α = (α-1) - α = -1<br/>α⁴ = -α<br/>α⁵ = -α²= -(α-1) = 1-α</p><p>Similarly: β⁵ = 1-β</p><p>α^2009 + β^2009 = α⁵ + β⁵ = (1-α) + (1-β) = 2 - (α+β) = 2 - 1 = <strong>1</strong></p><p>∴ Answer: C (which equals 1)</p>
Correct Answer: C