<p>Sum of common roots of the equations \(z^3 + 2z^2 + 2z + 1 = 0\) and \(z^{1985} + z^{100} + 1 = 0\) is</p>
Step-by-Step Solution
Key Concept: Factor the cubic as (z+1)(z²+z+1)=0 to get roots {-1, ω, ω²} where ω is a primitive cube root of unity, then check which satisfy z^1985 + z^100 + 1 = 0 using properties ω³=1 and 1+ω+ω²=0.
<p><strong>Step 1:</strong> Factor the cubic equation: z³ + 2z² + 2z + 1 = (z+1)(z² + z + 1) = 0</p><p>Roots are: z = -1, ω, ω² where ω = e^(2πi/3) is a primitive cube root of unity</p><p><strong>Step 2:</strong> Check z = -1 in z^1985 + z^100 + 1 = 0:<br/>(-1)^1985 + (-1)^100 + 1 = -1 + 1 + 1 = 1 ≠ 0 ✗</p><p><strong>Step 3:</strong> Check z = ω where ω³ = 1 and 1 + ω + ω² = 0:<br/>1985 ≡ 2 (mod 3) and 100 ≡ 1 (mod 3)<br/>ω^1985 + ω^100 + 1 = ω² + ω + 1 = 0 ✓</p><p><strong>Step 4:</strong> Check z = ω² where (ω²)³ = 1:<br/>1985 ≡ 2 (mod 3) and 100 ≡ 1 (mod 3)<br/>(ω²)^1985 + (ω²)^100 + 1 = ω⁴ + ω² + 1 = ω + ω² + 1 = 0 ✓</p><p><strong>Step 5:</strong> Sum of common roots = ω + ω² = -1 (using 1 + ω + ω² = 0)</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: A