Okay, so I have this problem where ω is a complex number satisfying ω³ = 1. I need to find all possible values of the expression 1/(1 + ω) + 1/(1 + ω²). Hmm, let me start by recalling that ω being a cube root of unity. The cube roots of unity are 1, ω, and ω², where ω ≠ 1. Also, they satisfy the equation ω² + ω + 1 = 0. That might come in handy.

First, maybe I should figure out what ω and ω² are. Since ω³ = 1, ω is a primitive cube root of unity, right? So ω ≠ 1, and ω² is the other primitive root. Also, ω² is the complex conjugate of ω if we consider ω as e^(2πi/3). But maybe I don't need to get into that much detail.

The expression is 1/(1 + ω) + 1/(1 + ω²). Let me try to simplify this. Maybe combine the fractions? Let's see.

First, let's compute the denominators. If I add 1 and ω, and 1 and ω², then the sum would be 1/(1 + ω) + 1/(1 + ω²). To add these two fractions, I need a common denominator. The denominators are (1 + ω) and (1 + ω²), so the common denominator would be (1 + ω)(1 + ω²). Let's compute that.

First, multiply the numerators: 1*(1 + ω²) + 1*(1 + ω) = (1 + ω²) + (1 + ω) = 2 + ω + ω². Then the denominator is (1 + ω)(1 + ω²). Let's compute that:

(1 + ω)(1 + ω²) = 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. Wait a minute, that's the same as the numerator! So the numerator is 2 + ω + ω², and the denominator is also 2 + ω + ω². Therefore, the entire expression simplifies to 1. Wait, that's interesting. So the sum is 1? But hold on, is that possible?

Wait, let me check that again. Let's compute the numerator and denominator step by step.

Numerator:

1/(1 + ω) + 1/(1 + ω²) = [ (1 + ω²) + (1 + ω) ] / (1 + ω)(1 + ω²) = (2 + ω + ω²) / (1 + ω + ω² + ω³).

But since ω³ = 1, the denominator becomes 1 + ω + ω² + 1 = 2 + ω + ω². Therefore, numerator and denominator are both 2 + ω + ω², so the ratio is 1. Therefore, the expression is equal to 1. So the only possible value is 1?

But wait, the problem says "find all possible values", which makes me think maybe there are multiple values depending on ω? But ω is a cube root of unity, and there are only two primitive ones: ω and ω². Let me check both cases.

Case 1: ω = 1. Then ω³ = 1, so ω = 1. Then the expression becomes 1/(1 + 1) + 1/(1 + 1²) = 1/2 + 1/2 = 1. So that's 1.

Case 2: ω is a primitive root, so ω ≠ 1. Then, as before, ω² + ω + 1 = 0. Therefore, the sum is 1. Wait, but in the problem statement, ω is a complex number such that ω³ = 1. So ω can be 1, ω, or ω². So if ω = 1, the value is 1. If ω is a primitive root, then the sum is 1. Therefore, the possible values are just 1?

But wait, hold on. Let me test with ω = 1. Then the expression is 1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 = 1. If ω is a primitive root, then ω² + ω + 1 = 0, so the expression is 1/(1 + ω) + 1/(1 + ω²). Let's compute this with ω² = -1 - ω. So, 1/(1 + ω) + 1/(1 + ω²) = 1/(1 + ω) + 1/(1 + (-1 - ω)) = 1/(1 + ω) + 1/(-ω). So that's 1/(1 + ω) - 1/ω. Let's compute that.

First, 1/(1 + ω) - 1/ω. Let's write both terms with a common denominator. The common denominator is ω(1 + ω). So:

[ω - (1 + ω)] / [ω(1 + ω)] = [ω - 1 - ω] / [ω(1 + ω)] = (-1)/[ω(1 + ω)].

But ω(1 + ω) = ω + ω². Since ω² = -1 - ω, this is ω -1 - ω = -1. Therefore, the denominator is -1, so the expression is (-1)/(-1) = 1. So even in this case, the sum is 1. Therefore, regardless of whether ω is 1 or a primitive cube root of unity, the expression is 1. Therefore, the only possible value is 1. So the answer is 1.

But wait, let me check again. Let's take ω = 1. Then the expression is 1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 = 1. If ω is a primitive root, say ω = e^(2πi/3) = -1/2 + i√3/2. Then ω² = e^(4πi/3) = -1/2 - i√3/2. Then 1 + ω = 1 + (-1/2 + i√3/2) = 1/2 + i√3/2. The reciprocal is 1/(1 + ω) = 1/(1/2 + i√3/2) = multiply numerator and denominator by 2: 2/(1 + i√3). Multiply numerator and denominator by the conjugate: (2)(1 - i√3)/(1 + 3) = (2)(1 - i√3)/4 = (1 - i√3)/2. Similarly, 1 + ω² = 1 + (-1/2 - i√3/2) = 1/2 - i√3/2. The reciprocal is 1/(1 + ω²) = 1/(1/2 - i√3/2) = multiply by 2: 2/(1 - i√3). Multiply by conjugate: (2)(1 + i√3)/(1 + 3) = (2)(1 + i√3)/4 = (1 + i√3)/2. Now add them: (1 - i√3)/2 + (1 + i√3)/2 = (1 - i√3 + 1 + i√3)/2 = 2/2 = 1. So even with ω as a primitive root, the sum is 1. Therefore, regardless of which cube root of unity ω is, the expression is 1. Therefore, the only possible value is 1.

Therefore, the answer is 1. So the possible values are just 1. So the final answer is \boxed{1}.

**Final Answer**
\boxed{1}
Thus, the possible values of the expression are $\boxed{1}$.
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