<Question>
Let $\omega$ be a complex number such that $\omega^3 = 1.$  Find all possible values of
\[\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}.\]Enter all the possible values, separated by commas.
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem. So, we have a complex number ω such that ω³ = 1. We need to find all possible values of the expression 1/(1 + ω) + 1/(1 + ω²). Hmm, alright. Let's start by recalling what the solutions to ω³ = 1 are.

First, the equation ω³ = 1 has three solutions, right? They are the cube roots of unity.

<prompt_1.1> Identify the cube roots of unity and their properties as outlined in the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
These are 1, ω, and ω², where ω ≠ 1. So, ω can be either of the non-real cube roots of unity. I remember that the cube roots of unity satisfy the equation ω² + ω + 1 = 0. That might come in handy later.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
The problem asks for all possible values of the expression 1/(1 + ω) + 1/(1 + ω²). Since ω can be 1, ω, or ω², but ω ≠ 1 because if ω were 1, then ω³ would be 1, but the other roots are the non-real ones. Wait, actually, ω could be 1, but then ω² would also be 1, so maybe we need to consider all possible ω satisfying ω³ = 1, including ω = 1. Let me check that.

<prompt_2.1> Consider the case when ω = 1 and then when ω is a non-real cube root of unity, as outlined in the above plan. </prompt_2.1>
-----

But let's check if there's another way to approach this, maybe by combining the fractions. Let's try adding the two fractions:

<prompt_2.2> Explore an alternative method by combining the fractions as outlined in the above plan. </prompt_2.2>

Alternatively, maybe rationalizing each term. Let's try that.

<prompt_2.3> Rationalize each term of the expression as outlined in the above plan. </prompt_2.3>

Alternatively, another approach: Let's denote S = 1/(1 + ω) + 1/(1 + ω²). Let's compute S.

<prompt_2.4> Consider another approach to compute the expression and verify consistency as outlined in the above plan. </prompt_2.4>

Wait, but let me check if there's a case when ω = 1. If ω = 1, then the original expression becomes 1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 = 1.
However, if ω = 1, then ω³ = 1 is satisfied, so ω = 1 is a valid root. But earlier, when I considered ω = 1, substituting into ω² + ω + 1 = 0, we get 1 + 1 + 1 = 3 ≠ 0, so in that case, the denominator 1 + ω = 2, which is non-zero, so the expression is defined.

<prompt_2.5> Re-evaluate the case when ω = 1 and confirm the results as outlined in the above plan. </prompt_2.5>
</Plan_2>

<Execution_2>
<execution_2.1>
If ω is a cube root of unity, then ω³ = 1. The roots are 1, ω, and ω². So, ω can be 1, or the primitive roots ω and ω². But if ω is 1, then ω² is also 1. So, substituting ω = 1 into the expression, we get 1/(1 + 1) + 1/(1 + 1) = (1/2) + (1/2) = 1. So, that's one possible value.

But wait, maybe the problem considers ω as a primitive cube root of unity, i.e., ω ≠ 1. Let me check the problem statement again. It says "Let ω be a complex number such that ω³ = 1". So, ω can be any of the three cube roots of unity, including 1. Therefore, we need to consider all three cases: ω = 1, ω = ω, and ω = ω². But when ω = ω, we have the primitive roots. Wait, but ω and ω² are distinct roots. So, perhaps there are two distinct cases: ω = 1 and ω ≠ 1. Let's explore that.
</execution_2.1>

<execution_2.2>
The expression is 1/(1 + ω) + 1/(1 + ω²). To add these, we need a common denominator, which would be (1 + ω)(1 + ω²). Let's compute the numerator:

[ (1 + ω²) + (1 + ω) ] all over (1 + ω)(1 + ω²). So the numerator becomes 1 + ω² + 1 + ω = 2 + ω + ω². 

Hmm, and the denominator is (1 + ω)(1 + ω²). Let's expand that: 1*1 + 1*ω² + ω*1 + ω*ω² = 1 + ω² + ω + ω³. But since ω³ = 1, this becomes 1 + ω² + ω + 1 = 2 + ω + ω². 

So the numerator is 2 + ω + ω² and the denominator is also 2 + ω + ω². Therefore, the entire expression simplifies to (2 + ω + ω²)/(2 + ω + ω²) = 1. Wait, so does that mean the value is 1?
</execution_2.2>

<execution_2.3>
The denominators are 1 + ω and 1 + ω². To add them, we need a common denominator, which would be (1 + ω)(1 + ω²). So:

[ (1 + ω²) + (1 + ω) ] / [ (1 + ω)(1 + ω²) ]

Simplify the numerator: 1 + ω² + 1 + ω = 2 + ω + ω². But from the equation ω² + ω + 1 = 0, we know that ω + ω² = -1. Therefore, the numerator becomes 2 + (-1) = 1.

So the numerator is 1. Now the denominator: (1 + ω)(1 + ω²). Let's expand this:

1*(1) + 1*ω² + ω*1 + ω*ω² = 1 + ω² + ω + ω³.

But ω³ = 1, so this becomes 1 + ω² + ω + 1 = 2 + ω + ω². Again, ω + ω² = -1, so denominator is 2 + (-1) = 1.

Therefore, the entire expression simplifies to 1/1 = 1.
</execution_2.3>

<execution_2.4>
First, combine the fractions: S = [ (1 + ω²) + (1 + ω) ] / [ (1 + ω)(1 + ω²) ].

Simplify the numerator: (1 + ω² + 1 + ω) = 2 + ω + ω².

From the equation ω² + ω + 1 = 0, we know that ω + ω² = -1. Therefore, the numerator becomes 2 + (-1) = 1.

So, S = 1 / [ (1 + ω)(1 + ω²) ].

Now, let's compute the denominator: (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³.

But since ω³ = 1, this becomes 1 + ω + ω² + 1 = (1 + 1) + (ω + ω²) = 2 + (-1) = 1.

Therefore, S = 1 / 1 = 1.
</execution_2.4>

<execution_2.5>
But when ω ≠ 1, then ω and ω² are the non-real roots satisfying ω² + ω + 1 = 0. So, in that case, the expression evaluates to 0. Therefore, the possible values are 0 and 1.
</execution_2.5>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let's verify:

Case 1: ω = 1. Then, ω² = 1. So, substituting into the expression:

<prompt_3.1> Verify the expression when ω = 1 as outlined in the above plan. </prompt_3.1>
-----

Based on execution_2.2, execution_2.3, execution_2.4:
So, regardless of the value of ω (as long as ω³ = 1 and the denominators are non-zero), the expression equals 1. Wait, but when ω = 1, the denominator 1 + ω = 2, which is fine, and 1 + ω² = 2 as well. So, when ω = 1, the expression is 1/2 + 1/2 = 1. But in the case when ω ≠ 1, we saw that the expression is 0. Wait, no, in the case when ω ≠ 1, the expression evaluates to 1 as per the algebraic manipulation.

<prompt_3.2> Re-evaluate the algebraic approach and confirm the results as outlined in the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 = 1. So, that's one possible value: 1.

Case 2: ω ≠ 1, so ω is a primitive cube root of unity. Then, ω² + ω + 1 = 0. Therefore, ω + ω² = -1. Let's see if we can use that.

So, the expression is 1/(1 + ω) + 1/(1 + ω²). Let's compute each term. Maybe we can combine them. Let me try to compute this sum.
</execution_3.1>

<execution_3.2>
Wait, that seems conflicting with my previous thought. Let me check again. If ω ≠ 1, then ω and ω² are roots of ω² + ω + 1 = 0. Therefore, ω + ω² = -1 and ω³ = 1. Then, when we compute the expression:

1/(1 + ω) + 1/(1 + ω²) = [ (1 + ω²) + (1 + ω) ] / [ (1 + ω)(1 + ω²) ] = (2 + ω + ω²) / (2 + ω + ω²) = 1. So, regardless of ω being 1 or not, as long as ω³ = 1, the expression equals 1? But when ω = 1, we have 1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 = 1. So, even when ω = 1, the expression is 1. Then, in both cases, the expression is 1. So, the only possible value is 1?
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
But that contradicts my earlier thought that when ω ≠ 1, the expression is 0. Wait, no, perhaps I made a mistake there. Let me re-examine.

<prompt_4.1> Confirm the results and understand the algebraic manipulation as outlined in the above plan. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Wait, when I combined the fractions, the numerator and denominator both become 2 + ω + ω². But since ω² + ω + 1 = 0, then ω + ω² = -1, so 2 + ω + ω² = 2 - 1 = 1. Therefore, numerator is 1 and denominator is 1, so the expression is 1. Therefore, regardless of ω being 1 or a primitive root, the expression is 1.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But when I first considered ω ≠ 1, I thought that ω + ω² = -1, so substituting into the numerator 2 + ω + ω² gives 1, and the denominator similarly gives 1, so the expression is 1. But when ω = 1, substituting directly gives 1. So, in both cases, the expression is 1. Therefore, the only possible value is 1.

But wait, that seems to suggest that regardless of ω being 1 or a primitive cube root of unity, the expression is always 1. Therefore, the only possible value is 1. But that contradicts my initial thought that when ω ≠ 1, the expression is 0. Wait, where did I get that from? Let me check again.

<prompt_5.1> Re-confirm the results and finalize the conclusion as outlined in the above plan. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, maybe I confused this problem with another one. Let me verify again. Let's take a specific example. Let ω be a primitive cube root of unity, say ω = e^(2πi/3) = -1/2 + i√3/2. Then ω² = e^(4πi/3) = -1/2 - i√3/2. Let's compute 1/(1 + ω) + 1/(1 + ω²).

First, compute 1 + ω = 1 + (-1/2 + i√3/2) = 1/2 + i√3/2. The modulus squared is (1/2)^2 + (√3/2)^2 = 1/4 + 3/4 = 1, so modulus is 1. The inverse of 1 + ω is (1/2 - i√3/2)/1 = 1/2 - i√3/2. Similarly, 1 + ω² = 1 + (-1/2 - i√3/2) = 1/2 - i√3/2. The inverse is (1/2 + i√3/2)/1 = 1/2 + i√3/2. Adding these two inverses: (1/2 - i√3/2) + (1/2 + i√3/2) = 1. So, indeed, the sum is 1. So, even when ω is a primitive root, the expression equals 1. Therefore, regardless of ω being 1 or a primitive root, the expression is 1. Therefore, the only possible value is 1.

But wait, that seems to conflict with the initial thought that when ω ≠ 1, the sum is 0. But in reality, through computation and algebraic manipulation, it's clear that the expression simplifies to 1 in all cases. Therefore, the answer is 1.

Therefore, all possible values are just 1. So, the answer is \boxed{1}.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
**Final Answer**
\boxed{1}
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Therefore, the only possible value for the given expression is $\boxed{1}$.
</Final_answer>
</think>