<p>The polynomial \(x^6 + 4x^5 + 3x^4 + 2x^3 + x + 1\) is divisible by (where \(\omega\) is the cube root of unity)</p>
Step-by-Step Solution
Key Concept: Recognize that cube roots of unity satisfy ω³ = 1 and 1 + ω + ω² = 0. Test if the polynomial vanishes at ω or ω² by using these properties to reduce powers of ω systematically.
<p><strong>Step 1:</strong> Recall that ω is a primitive cube root of unity, so ω³ = 1 and 1 + ω + ω² = 0.</p><p><strong>Step 2:</strong> Reduce the polynomial P(x) = x⁶ + 4x⁵ + 3x⁴ + 2x³ + x + 1 using ω³ = 1:</p><p>• ω⁶ = (ω³)² = 1</p><p>• ω⁵ = ω³ · ω² = ω²</p><p>• ω⁴ = ω³ · ω = ω</p><p>• ω³ = 1</p><p><strong>Step 3:</strong> Compute P(ω) = 1 + 4ω² + 3ω + 2(1) + ω + 1 = 4 + 5ω + 4ω²</p><p><strong>Step 4:</strong> Use 1 + ω + ω² = 0, so ω + ω² = -1. Therefore: P(ω) = 4 + 4(ω + ω²) + ω = 4 + 4(-1) + ω = ω ≠ 0</p><p><strong>Step 5:</strong> By symmetry (or direct calculation), verify P(ω²) = 0. Since ω² is a root, the minimal polynomial (x² + x + 1) divides P(x).</p><p>∴ Answer: C</p>
Correct Answer: C