If a is a real number satisfying a^2 + 1 = a, compute the exact value of a^12 + a^6 + 1. Show the reasoning clearly and simplify powers using any useful identities.

Introduction / Context:
This algebraic simplification question assesses your ability to use polynomial identities and properties of special roots. The equation a^2 + 1 = a can be rearranged into a quadratic with complex roots. From that relation, we can deduce periodic behavior of powers of a and evaluate expressions such as a^12 + a^6 + 1 without expanding naively.

Given Data / Assumptions:

• a is a real or complex number satisfying a^2 + 1 = a.
• We must find the exact value of a^12 + a^6 + 1.
• Standard algebraic identities and power-cycling techniques may be used.

Concept / Approach:
Start by forming the quadratic: a^2 − a + 1 = 0. The roots of this quadratic are complex and lie on the unit circle. Numbers on the unit circle often have powers that repeat in cycles. In particular, the roots are primitive 6th roots of unity excluding 1, which implies a^6 = 1. Once a^6 is known, higher powers like a^12 follow immediately, enabling a quick evaluation of the target sum.

Step-by-Step Solution:
From a^2 + 1 = a, rearrange to a^2 − a + 1 = 0.This quadratic has roots a = (1 ± i√3)/2, which satisfy a^6 = 1.Therefore a^6 = 1 and a^12 = (a^6)^2 = 1.Compute the sum: a^12 + a^6 + 1 = 1 + 1 + 1 = 3.

Verification / Alternative check:
Another route is to note a^2 = a − 1, then power-reduce systematically: a^3 = a·a^2 = a(a − 1) = a^2 − a = (a − 1) − a = −1. Hence a^6 = (a^3)^2 = (−1)^2 = 1. The same conclusion leads to a^12 = 1 and the sum equals 3.

Why Other Options Are Wrong:

• 2 and 1: These ignore that both a^6 and a^12 equal 1, causing undercount.
• −3 and −1: Sign errors from misusing the relation a^2 = a − 1 or mis-evaluating a^3.

Common Pitfalls:
Expanding powers directly without using the quadratic relation; forgetting that the roots are 6th roots of unity; arithmetic slips in deriving a^3.

Final Answer:
3