<p>If \(x^2 + x + 1 = 0\), then the value of \(\left(x + \dfrac{1}{x}\right)^2 + \left(x^2 + \dfrac{1}{x^2}\right)^2 + \cdots + \left(x^{27} + \dfrac{1}{x^{27}}\right)^2\) is</p>
Step-by-Step Solution
Key Concept: Recognize that x² + x + 1 = 0 gives x = ω or ω² (cube roots of unity), so x³ = 1 and the sum has a periodic pattern with period 3. The terms xⁿ + 1/xⁿ depend only on n mod 3.
<p><strong>Step 1:</strong> From x² + x + 1 = 0, we have x³ = 1 (roots are ω and ω² where ω = e^(2πi/3)). Also, 1/x = x².</p><p><strong>Step 2:</strong> Find the repeating values:</p><p>• For n ≡ 1 (mod 3): x + 1/x = x + x² = −1 (from x² + x + 1 = 0)</p><p>So (x + 1/x)² = 1</p><p>• For n ≡ 2 (mod 3): x² + 1/x² = x² + x⁴ = x² + x = −1</p><p>So (x² + 1/x²)² = 1</p><p>• For n ≡ 0 (mod 3): x³ + 1/x³ = 1 + 1 = 2</p><p>So (x³ + 1/x³)² = 4</p><p><strong>Step 3:</strong> Among n = 1, 2, ..., 27:</p><p>• n ≡ 1 (mod 3): n ∈ {1, 4, 7, ..., 25} → 9 terms, each contributing 1</p><p>• n ≡ 2 (mod 3): n ∈ {2, 5, 8, ..., 26} → 9 terms, each contributing 1</p><p>• n ≡ 0 (mod 3): n ∈ {3, 6, 9, ..., 27} → 9 terms, each contributing 4</p><p><strong>Step 4:</strong> Total = 9(1) + 9(1) + 9(4) = 9 + 9 + 36 = <strong>54</strong></p><p>∴ Answer: D</p>
Correct Answer: D