<p>If \(\alpha^{14}+\alpha^{10}+\alpha^6+\alpha^2+1=0\), then \(\alpha\) can be:</p>
Step-by-Step Solution
Key Concept: Factor: \alpha^1^4+\alpha^1^0+\alpha^6+\alpha^2+1 = \alpha^2(\alpha^1^2 + \alpha^8 + \alpha^4 + \alpha^0) + 1. This is related to cyclotomic polynomials. Dividing by \alpha^7: the equation becomes symmetric in nature.
<p>The equation $\sum_{k=0}^4 \alpha^{4k+2}=0$ is satisfied by 10th or 14th roots of unity. Testing $\alpha=e^{\pi i/7}=e^{2\pi i/14}$: this is a primitive 14th root, and by properties of cyclotomic polynomials, it satisfies the equation. ✓</p>
Correct Answer: C