<p>Find the common roots of \(x^{12} - 1 = 0\) and \(x^4 + x^2 + 1 = 0\).</p>
Step-by-Step Solution
Key Concept: Recognize that x⁴ + x² + 1 factors as (x² - ω)(x² - ω²) where ω = e^(2πi/3), and these factors correspond to 6th roots of unity that are not cube roots of unity.
<p><strong>Step 1:</strong> Factor x⁴ + x² + 1. Multiply by (x² - 1):</p><p>(x² - 1)(x⁴ + x² + 1) = x⁶ - 1</p><p>So x⁴ + x² + 1 = (x⁶ - 1)/(x² - 1), meaning roots of x⁴ + x² + 1 are 6th roots of unity except ±1.</p><p><strong>Step 2:</strong> The 12th roots of unity are e^(2πik/12) for k = 0,1,...,11. The 6th roots of unity are e^(2πik/6) for k = 0,1,...,5.</p><p><strong>Step 3:</strong> Roots of x⁴ + x² + 1 = 0 are the 6th roots of unity except ±1, which are: e^(2πi/6) = e^(πi/3), e^(4πi/6) = e^(2πi/3), e^(8πi/6) = e^(4πi/3), e^(10πi/6) = e^(5πi/3)</p><p><strong>Step 4:</strong> These can be written as ω, ω², ω⁴, ω⁵ where ω = e^(2πi/6). Note ω³ = -1, so ω⁴ = -ω and ω⁵ = -ω².</p><p><strong>Step 5:</strong> All these roots (ω, ω², -ω, -ω²) are also 12th roots of unity.</p><p>∴ Answer: x = ±ω, ±ω² (where ω = e^(πi/3))</p>
Correct Answer: x = ±ω², ±ω