Question:

<p><strong>Step 1: Recall properties of cube roots of unity.</strong> If ω is a cube root of unity (ω ≠ 1), then:</p><ul><li>ω³ = 1</li><li>1 + ω + ω² = 0, which implies ω + ω² = -1</li></ul><p><strong>Step 2: Simplify x = ω - ω² - 2.</strong></p><p>We need to express this in terms of known quantities. Let ω = e^(2πi/3), so ω² = e^(4πi/3).</p><p>However, a more direct approach: Since we only know ω + ω² = -1, let's compute ω - ω² separately.</p><p>Note that (ω - ω²)² = ω² - 2ω·ω² + ω⁴ = ω² - 2ω³·ω + ω·ω³ = ω² - 2ω + ω = ω² - ω.</p><p>Actually, let's use: ω - ω² = ω - ω² and ω + ω² = -1.</p><p>From these: (ω - ω²)² = (ω + ω²)² - 4ωω² = (-1)² - 4ω³ = 1 - 4(1) = -3.</p><p>So ω - ω² = ±i√3.</p><p><strong>Step 3: Determine the value of x.</strong></p><p>For ω = e^(2πi/3) = -1/2 + i√3/2, we have ω² = e^(4πi/3) = -1/2 - i√3/2.</p><p>Thus: ω - ω² = (i√3/2) - (-i√3/2) = i√3.</p><p>Therefore: x = i√3 - 2.</p><p><strong>Step 4: Calculate x + 2 = i√3, so (x+2)² = -3, giving x² + 4x + 4 = -3, thus x² + 4x + 7 = 0.</strong></p><p><strong>Step 5: Use the relation x² = -4x - 7 to reduce higher powers.</strong></p><p>x³ = x·x² = x(-4x - 7) = -4x² - 7x = -4(-4x - 7) - 7x = 16x + 28 - 7x = 9x + 28.</p><p>x⁴ = x·x³ = x(9x + 28) = 9x² + 28x = 9(-4x - 7) + 28x = -36x - 63 + 28x = -8x - 63.</p><p><strong>Step 6: Substitute into the polynomial.</strong></p><p>x⁴ + 3x³ + 2x² - 11x - 6</p><p>= (-8x - 63) + 3(9x + 28) + 2(-4x - 7) - 11x - 6</p><p>= -8x - 63 + 27x + 84 - 8x - 14 - 11x - 6</p><p>= (-8 + 27 - 8 - 11)x + (-63 + 84 - 14 - 6)</p><p>= 0·x + 1</p><p>= 1.</p><p><strong>∴ Answer: 1</strong></p>