<p>Find the sum of squares of all roots of the equation \(x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 = 0\).</p>
Step-by-Step Solution
Key Concept: Recognize this polynomial as a geometric series sum: (x^9 + 1)/(x + 1) for x ≠ -1. Use Vieta's formulas on the resulting roots to find Σr_i² = (Σr_i)² - 2Σr_i·r_j.
<p><strong>Step 1:</strong> Recognize the polynomial as a geometric series. Multiply by (x+1):</p><p>(x+1)(x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) = x^9 + 1</p><p>So our polynomial equals (x^9 + 1)/(x + 1) for x ≠ -1.</p><p><strong>Step 2:</strong> The roots are the 9th roots of -1 except x = -1. These are: x = e^(iπ(2k+1)/9) for k = 0,1,2,...,8, excluding k = 4 (which gives -1).</p><p><strong>Step 3:</strong> Apply Vieta's formulas to x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 = 0:</p><p>• Sum of roots: Σr_i = 1 (coefficient of x^7 with sign change)</p><p>• Sum of products of pairs: Σr_i·r_j = 1 (coefficient of x^6)</p><p><strong>Step 4:</strong> Use the identity: (Σr_i)² = Σr_i² + 2Σr_i·r_j</p><p>Therefore: Σr_i² = (Σr_i)² - 2Σr_i·r_j = (1)² - 2(1) = 1 - 2 = -1</p><p>∴ Answer: -1</p>
Correct Answer: -1