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.
Aptitude
Simplification
Difficulty: Easy
Choose an option
Answer
Correct Answer: 3
Explanation
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:
a^3 = 1 ⇒ a^6 = (a^3)^2 = 1a^12 = (a^3)^4 = 1Therefore a^12 + a^6 + 1 = 1 + 1 + 1 = 3Verification / Alternative check:
From a^2 + a + 1 = 0 ⇒ a^2 = -a - 1. Iteratively computing powers confirms the period-3 cycle.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:
3