<p>Given <span>\(\alpha, \beta\)</span>, respectively, the fifth and the fourth non-real roots of unity, respectively, then find the value of <span>\((1 + \alpha)(1 + \beta)(1 + \alpha^2)(1 + \beta^2)(1 + \alpha^4)(1 + \beta^4)\)</span>.</p>
Step-by-Step Solution
Key Concept: Use the factorization of cyclotomic polynomials: for nth roots of unity, ∏(1 + ω^k) can be evaluated using the identity that ∏(x - ω^k) relates to cyclotomic polynomials. For α (5th root) and β (4th root), recognize that (1 + α)(1 + α²)(1 + α⁴) and (1 + β)(1 + β²)(1 + β⁴) can be computed separately using the property that these products often yield 0 or 1.
<p><strong>Step 1:</strong> Identify the roots. Let α be a primitive 5th root of unity (α⁵ = 1, α ≠ 1) and β be a primitive 4th root of unity (β⁴ = 1, β ≠ 1).</p><p><strong>Step 2:</strong> For a primitive 4th root of unity β, we have β⁴ = 1. The fourth roots of unity are {1, i, -1, -i}. For the non-real primitive 4th root, we can take β = i or β = -i.</p><p><strong>Step 3:</strong> If β = i, then β² = i² = -1. Therefore, (1 + β²) = (1 + (-1)) = 0.</p><p><strong>Step 4:</strong> Since one factor in the product is zero: (1 + α)(1 + β)(1 + α²)(1 + β²)(1 + α⁴)(1 + β⁴) = (1 + α)(1 + β)(1 + α²) · 0 · (1 + α⁴)(1 + β⁴) = 0.</p><p><strong>Step 5:</strong> Similarly, if β = -i, then β² = (-i)² = -1, yielding the same result.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0