Roots of unity style evaluation (corrected): If a satisfies a^2 + a + 1 = 0 (i.e., a is a non-real cube root of unity), compute the value of a^12 + a^6 + 1.

Introduction / Context:This problem uses properties of complex cube roots of unity. If a is a non-real cube root of unity, then a^3 = 1, a ≠ 1, and a^2 + a + 1 = 0. Powers of a cycle with period 3, enabling quick evaluation of higher powers.

Given Data / Assumptions:

• a^2 + a + 1 = 0
• a^3 = 1 and a ≠ 1

Concept / Approach:Use the periodicity: a^k repeats every 3 steps. Specifically, a^0 = 1, a^1 = a, a^2 = a^2, a^3 = 1, a^4 = a, etc. Reduce exponents modulo 3 to simplify a^12 and a^6.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 2/1: Would require one of a^6 or a^12 to be 0, which never occurs for cube roots of unity.
• -3: Sign error; all three terms equal 1, summing to +3.

Common Pitfalls:

• Misinterpreting the condition or forgetting that a ≠ 1.
• Not reducing exponents modulo 3 before evaluating.

Final Answer: