Explanation:
Okay, so Jim and Martha are at the corner of a rectangular field. The field is 300 feet wide and 400 feet long. Jim walks diagonally across the field, while Martha walks along the length and width to get to the same location. The question is asking how many feet less Jim walks compared to Martha.

First, let me visualize the problem. There's a rectangle, right? Let me sketch it mentally. Let's say the rectangle has a width of 300 feet and a length of 400 feet. So, if Jim and Martha are at one corner, let's say the bottom-left corner, then Martha would walk along the length (400 feet) and then along the width (300 feet) to reach the opposite corner. Wait, but actually, if they are at the corner, and Martha is walking along the length and width, does that mean she walks along both the length and the width? So, Martha's total distance would be the sum of the length and the width? Let me confirm.

Yes, Martha is walking along the length and the width. So, if the field is 400 feet long and 300 feet wide, then Martha would walk 400 feet along the length and then 300 feet along the width, making her total distance 400 + 300 = 700 feet.

But Jim is taking the diagonal path. So, instead of going along the length and width, he goes straight across the field diagonally. The question is asking how many feet less Jim walks compared to Martha. So, we need to find the difference between Martha's distance and Jim's distance.

To find Jim's distance, we need to calculate the length of the diagonal of the rectangle. Since the field is a rectangle, the diagonal can be found using the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this case, the diagonal of the rectangle forms a right-angled triangle with the length and the width of the rectangle. So, if we consider the length as one side (400 feet) and the width as the other side (300 feet), then the diagonal (which is Jim's path) is the hypotenuse.

So, let me write that down. Let me denote the length as L = 400 feet, the width as W = 300 feet, and the diagonal as D. Then according to the Pythagorean theorem:

D² = L² + W²

So, plugging in the numbers:

D² = 400² + 300²

Let me compute that. First, 400 squared is 160,000. Then, 300 squared is 90,000. Adding those together gives 160,000 + 90,000 = 250,000.

Therefore, D² = 250,000. To find D, we take the square root of 250,000.

What's the square root of 250,000? Well, sqrt(250,000) can be simplified. Let me think. 250,000 is 25 * 10,000. The square root of 25 is 5, and the square root of 10,000 is 100. Therefore, sqrt(25 * 10,000) = sqrt(25) * sqrt(10,000) = 5 * 100 = 500.

So, the diagonal D is 500 feet.

Therefore, Jim walks 500 feet, while Martha walks 700 feet. The question is asking how many feet less Jim walks than Martha. So, we subtract Jim's distance from Martha's distance: 700 - 500 = 200 feet.

Wait, so Jim walks 200 feet less than Martha? Let me check my calculations again to make sure I didn't make a mistake.

First, Martha's path: 400 + 300 = 700 feet. That seems straightforward.

Jim's path: diagonal. Using Pythagoras, 400² + 300² = 160,000 + 90,000 = 250,000. Square root of 250,000 is indeed 500. So, 500 feet.

Difference: 700 - 500 = 200. So, 200 feet less.

But let me think again. Is there a possibility that Martha is walking along the length and width, but maybe not both? Wait, the problem says "walking along its length and width." So, if they start at one corner, to get to the opposite corner, Martha would have to walk along both the length and the width. So, that's 400 + 300.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's not going to the opposite corner? Wait, the problem says "the same location." So, if Jim is walking diagonally across the field, then Martha is going to the same location, which is the opposite corner. Therefore, Martha's path is along the length and the width.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's not going all the way? Wait, no. If they start at the same corner, and they end at the same location, which is the opposite corner. So, Martha would have to walk along the length and then the width, or along the width and then the length. Either way, the total distance is 400 + 300 = 700.

So, that seems correct.

Alternatively, maybe the problem is referring to something else? Let me check the problem statement again.

"Jim and Martha are standing together at the corner of a rectangular field. Jim walks diagonally across the field. Martha gets to the same location by walking along its length and width. The field is 300 feet wide and 400 feet long. How many feet less than Martha does Jim walk?"

So, "walking along its length and width." So, Martha walks along the length and the width. So, that would be the two sides of the rectangle. So, total distance is length + width.

Therefore, Martha's distance is 400 + 300 = 700. Jim's distance is 500. So, difference is 200.

Therefore, the answer is 200 feet.

But let me think again. Maybe there's a trick here? For example, maybe Martha is walking along the length and then the width, but maybe she's not going to the opposite corner? Wait, but the problem says "the same location," which is the opposite corner.

Alternatively, maybe the problem is referring to Martha walking along the length and then the width, but maybe she's not going all the way? But that seems unlikely.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along both the length and the width, but not necessarily the entire length and width? But that would be ambiguous.

But given the problem says "the same location," which is the opposite corner, so Martha must walk the entire length and the entire width.

Alternatively, maybe the problem is referring to Martha walking along the length and then the width, but maybe she's taking a different path? But no, the problem says "walking along its length and width," which would imply the two sides.

Alternatively, maybe the problem is in the terminology. For example, sometimes "length" and "width" can be confusing. But the problem says the field is 300 feet wide and 400 feet long. So, width is 300, length is 400. So, if they are at the corner, then the two sides adjacent to that corner are the length and the width. So, Martha walks along those two sides.

Therefore, Martha's total distance is 400 + 300 = 700.

Jim walks the diagonal, which is 500.

So, 700 - 500 = 200.

Therefore, Jim walks 200 feet less than Martha.

But let me check once more. Maybe I made a mistake in the Pythagorean theorem? Let me recalculate.

400 squared is 400*400. Let me compute that. 400*400 is 160,000. 300 squared is 90,000. Adding them gives 250,000. Square root of 250,000.

Well, sqrt(250,000) = sqrt(25 * 10,000) = 5 * 100 = 500. Correct.

Alternatively, sqrt(250,000) can be thought of as sqrt(250 * 1000). Wait, but that might not help. Alternatively, sqrt(250,000) = sqrt(25 * 10,000) = 5 * 100 = 500.

Yes, that's correct.

Alternatively, maybe the problem is in units? The field is 300 feet wide and 400 feet long. So, all units are in feet. So, the answer is 200 feet.

Alternatively, maybe the problem is asking how many feet less than Martha does Jim walk? So, Martha walks more, Jim walks less. So, the difference is Martha's distance minus Jim's distance, which is 700 - 500 = 200.

Therefore, the answer is 200.

But let me check if there's another way to interpret the problem. For example, maybe Martha walks along the length and the width, but not necessarily both? For example, if the field is 300 feet wide and 400 feet long, maybe Martha walks along the length (400 feet) and then along the width (300 feet), but maybe she's not going to the opposite corner? Wait, but the problem says "the same location," so if they start at the corner, then the same location would be the opposite corner.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's walking along the length and then the width from a different starting point? But the problem says they are standing together at the corner.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along the length and then the width, but maybe she's not going all the way? But that would be unclear.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along the length and then the width, but maybe she's taking a different path? But no, the problem says "the same location," so it's the opposite corner.

Alternatively, maybe the problem is in the term "diagonally across the field." If the field is a rectangle, then the diagonal is from one corner to the opposite corner. So, that's what Jim is doing.

Therefore, I think my original calculation is correct.

So, Martha walks 700 feet, Jim walks 500 feet, so Jim walks 200 feet less.

Therefore, the answer is 200.

But just to make sure, let me think of another way. Suppose the field is 300 feet wide and 400 feet long. So, imagine a rectangle with sides 300 and 400. The diagonal is 500. So, 3-4-5 triangle. Oh! Wait, 300, 400, 500 is a scaled-up 3-4-5 triangle. Because 3*100=300, 4*100=400, 5*100=500. So, that's a classic Pythagorean triple. So, that's why the diagonal is 500.

Therefore, that confirms that the diagonal is indeed 500.

Therefore, Martha walks 400 + 300 = 700.

Difference is 700 - 500 = 200.

Therefore, the answer is 200.

I think that's solid.

**Final Answer**
Jim walks \boxed{200} feet less than Martha.

To determine how many feet less Jim walks compared to Martha, we analyze the two different paths they take across a rectangular field.

---

### **Understanding the Problem**

- The field is a **rectangle** with:
  - **Width** = 300 feet
  - **Length** = 400 feet
- **Jim** walks **diagonally** across the field.
- **Martha** walks **along the length and width** to reach the same destination.
- We are asked to find how many **feet less** Jim walks than Martha.

---

### **Step 1: Calculate Martha's Distance**

Martha walks along the **length** and the **width** of the field. Since the field is a rectangle, she must walk the full length and the full width to reach the opposite corner.

$$
\text{Martha's distance} = \text{Length} + \text{Width} = 400 + 300 = 700 \text{ feet}
$$

---

### **Step 2: Calculate Jim's Distance**

Jim walks **diagonally** across the field. This diagonal forms the **hypotenuse** of a right triangle with the length and width as the other two sides. We can use the **Pythagorean Theorem** to find the length of the diagonal:

$$
D^2 = L^2 + W^2
$$
$$
D^2 = 400^2 + 300^2 = 160000 + 90000 = 250000
$$
$$
D = \sqrt{250000} = 500 \text{ feet}
$$

So, Jim walks **500 feet**.

---

### **Step 3: Find the Difference**

To find how many feet less Jim walks than Martha, subtract Jim's distance from Martha's:

$$
\text{Difference} = 700 - 500 = 200 \text{ feet}
$$

---

### **Final Answer**

$$
\boxed{200}
$$

[Assistant]
Explanation:
Okay, so Jim and Martha are at the corner of a rectangular field. The field is 300 feet wide and 400 feet long. Jim walks diagonally across the field, while Martha walks along the length and width to get to the same location. The question is asking how many feet less Jim walks compared to Martha.

First, let me visualize the problem. There's a rectangle, right? Let me sketch it mentally. Let's say the rectangle has a width of 300 feet and a length of 400 feet. So, if Jim and Martha are at one corner, let's say the bottom-left corner, then Martha would walk along the length (400 feet) and then along the width (300 feet) to reach the opposite corner. Wait, but actually, if they are at the corner, and Martha is walking along the length and width, does that mean she walks along both the length and the width? So, Martha's total distance would be the sum of the length and the width? Let me confirm.

Yes, Martha is walking along the length and the width. So, if the field is 400 feet long and 300 feet wide, then Martha would walk 400 feet along the length and then 300 feet along the width, making her total distance 400 + 300 = 700 feet.

But Jim is taking the diagonal path. So, instead of going along the length and width, he goes straight across the field diagonally. The question is asking how many feet less Jim walks compared to Martha. So, we need to find the difference between Martha's distance and Jim's distance.

To find Jim's distance, we need to calculate the length of the diagonal of the rectangle. Since the field is a rectangle, the diagonal can be found using the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this case, the diagonal of the rectangle forms a right-angled triangle with the length and the width of the rectangle. So, if we consider the length as one side (400 feet) and the width as the other side (300 feet), then the diagonal (which is Jim's path) is the hypotenuse.

So, let me write that down. Let me denote the length as L = 400 feet, the width as W = 300 feet, and the diagonal as D. Then according to the Pythagorean theorem:

D² = L² + W²

So, plugging in the numbers:

D² = 400² + 300²

Let me compute that. First, 400 squared is 160,000. Then, 300 squared is 90,000. Adding those together gives 160,000 + 90,000 = 250,000.

Therefore, D² = 250,000. To find D, we take the square root of 250,000.

What's the square root of 250,000? Well, sqrt(250,000) can be simplified. Let me think. 250,000 is 25 * 10,000. The square root of 25 is 5, and the square root of 10,000 is 100. Therefore, sqrt(25 * 10,000) = sqrt(25) * sqrt(10,000) = 5 * 100 = 500.

So, the diagonal D is 500 feet.

Therefore, Jim walks 500 feet, while Martha walks 700 feet. The question is asking how many feet less Jim walks than Martha. So, we subtract Jim's distance from Martha's distance: 700 - 500 = 200 feet.

Wait, so Jim walks 200 feet less than Martha? Let me check my calculations again to make sure I didn't make a mistake.

First, Martha's path: 400 + 300 = 700 feet. That seems straightforward.

Jim's path: diagonal. Using Pythagoras, 400² + 300² = 160,000 + 90,000 = 250,000. Square root of 250,000 is indeed 500. So, 500 feet.

Difference: 700 - 500 = 200. So, 200 feet less.

But let me think again. Is there a possibility that Martha is walking along the length and width, but maybe not both? Wait, the problem says "walking along its length and width." So, if they start at one corner, to get to the opposite corner, Martha would have to walk along both the length and the width. So, that's 400 + 300.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's not going to the opposite corner? Wait, the problem says "the same location." So, if Jim is walking diagonally across the field, then Martha is going to the same location, which is the opposite corner. Therefore, Martha's path is along the length and the width.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's not going all the way? Wait, no. If they start at the same corner, and they end at the same location, which is the opposite corner. So, Martha would have to walk along the length and then the width, or along the width and then the length. Either way, the total distance is 400 + 300 = 700.

So, that seems correct.

Alternatively, maybe the problem is referring to something else? Let me check the problem statement again.

"Jim and Martha are standing together at the corner of a rectangular field. Jim walks diagonally across the field. Martha gets to the same location by walking along its length and width. The field is 300 feet wide and 400 feet long. How many feet less than Martha does Jim walk?"

So, "walking along its length and width." So, Martha walks along the length and the width. So, that would be the two sides of the rectangle. So, total distance is length + width.

Therefore, Martha's distance is 400 + 300 = 700. Jim's distance is 500. So, difference is 200.

Therefore, the answer is 200 feet.

But let me think again. Maybe there's a trick here? For example, maybe Martha is walking along the length and then the width, but maybe she's not going to the opposite corner? Wait, but the problem says "the same location," which is the opposite corner.

Alternatively, maybe the problem is referring to Martha walking along the length and then the width, but maybe she's not going all the way? But that seems unlikely.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along both the length and the width, but not necessarily the entire length and width? But that would be ambiguous.

But given the problem says "the same location," which is the opposite corner, so Martha must walk the entire length and the entire width.

Alternatively, maybe the problem is referring to Martha walking along the length and then the width, but maybe she's taking a different path? But no, the problem says "walking along its length and width," which would imply the two sides.

Alternatively, maybe the problem is in the terminology. For example, sometimes "length" and "width" can be confusing. But the problem says the field is 300 feet wide and 400 feet long. So, width is 300, length is 400. So, if they are at the corner, then the two sides adjacent to that corner are the length and the width. So, Martha walks along those two sides.

Therefore, Martha's total distance is 400 + 300 = 700.

Jim walks the diagonal, which is 500.

So, 700 - 500 = 200.

Therefore, Jim walks 200 feet less than Martha.

But let me check once more. Maybe I made a mistake in the Pythagorean theorem? Let me recalculate.

400 squared is 400*400. Let me compute that. 400*400 is 160,000. 300 squared is 90,000. Adding them gives 250,000. Square root of 250,000.

Well, sqrt(250,000) = sqrt(25 * 10,000) = 5 * 100 = 500. Correct.

Alternatively, sqrt(250,000) can be thought of as sqrt(250 * 1000). Wait, but that might not help. Alternatively, sqrt(250,000) = sqrt(25 * 10,000) = 5 * 100 = 500.

Yes, that's correct.

Alternatively, maybe the problem is in units? The field is 300 feet wide and 400 feet long. So, all units are in feet. So, the answer is 200 feet.

Alternatively, maybe the problem is asking how many feet less than Martha does Jim walk? So, Martha walks more, Jim walks less. So, the difference is Martha's distance minus Jim's distance, which is 700 - 500 = 200.

Therefore, the answer is 200.

But let me check if there's another way to interpret the problem. For example, maybe Martha walks along the length and the width, but not necessarily both? For example, if the field is 300 feet wide and 400 feet long, maybe Martha walks along the length (400 feet) and then along the width (300 feet), but maybe she's not going to the opposite corner? Wait, but the problem says "the same location," so if they start at the corner, then the same location would be the opposite corner.

Alternatively, maybe Martha is walking along the length and then the width, but maybe she's walking along the length and then the width from a different starting point? But the problem says they are standing together at the corner.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along the length and then the width, but maybe she's not going all the way? But that would be unclear.

Alternatively, maybe the problem is referring to Martha walking along the length and the width, meaning that she walks along the length and then the width, but maybe she's taking a different path? But no, the problem says "the same location," so it's the opposite corner.

Alternatively, maybe the problem is in the term "diagonally across the field." If the field is a rectangle, then the diagonal is from one corner to the opposite corner. So, that's what Jim is doing.

Therefore, I think my original calculation is correct.

So, Martha walks 700 feet, Jim walks 500 feet, so Jim walks 200 feet less.

Therefore, the answer is 200.

But just to make sure, let me think of another way. Suppose the field is 300 feet wide and 400 feet long. So, imagine a rectangle with sides 300 and 400. The diagonal is 500. So, 3-4-5 triangle. Oh! Wait, 300, 400, 500 is a scaled-up 3-4-5 triangle. Because 3*100=300, 4*100=400, 5*100=500. So, that's a classic Pythagorean triple. So, that's why the diagonal is 500.

Therefore, that confirms that the diagonal is indeed 500.

Therefore, Martha walks 400 + 300 = 700.

Difference is 700 - 500 = 200.

Therefore, the answer is 200.

I think that's solid.

**Final Answer**
Jim walks \boxed{200} feet less than Martha.

To determine how many feet less Jim walks compared to Martha, we analyze the two different paths they take across a rectangular field.

---

### **Understanding the Problem**

- The field is a **rectangle** with:
  - **Width** = 300 feet
  - **Length** = 400 feet
- **Jim** walks **diagonally** across the field.
- **Martha** walks **along the length and width** to reach the same destination.
- We are asked to find how many **feet less** Jim walks than Martha.

---

### **Step 1: Calculate Martha's Distance**

Martha walks along the **length** and the **width** of the field. Since the field is a rectangle, she must walk the full length and the full width to reach the opposite corner.

$$
\text{Martha's distance} = \text{Length} + \text{Width} = 400 + 300 = 700 \text{ feet}
$$

---

### **Step 2: Calculate Jim's Distance**

Jim walks **diagonally** across the field. This diagonal forms the **hypotenuse** of a right triangle with the length and width as the other two sides. We can use the **Pythagorean Theorem** to find the length of the diagonal:

$$
D^2 = L^2 + W^2
$$
$$
D^2 = 400^2 + 300^2 = 160000 + 90000 = 250000
$$
$$
D = \sqrt{250000} = 500 \text{ feet}
$$

So, Jim walks **500 feet**.

---

### **Step 3: Find the Difference**

To find how many feet less Jim walks than Martha, subtract Jim's distance from Martha's:

$$
\text{Difference} = 700 - 500 = 200 \text{ feet}
$$

---

### **Final Answer**

$$
\boxed{200}
$$

Solution: The diagonal of the field is $\sqrt{300^2+400^2}=500$ feet long so Jim walks 500 feet. Two adjacent sides of the field are $300+400=700$ feet long so Martha walks 700 feet. Jim walks $700-500=\boxed{200}$ feet less than Martha.


Explanation:
Okay, so I need to find all possible values of the expression $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$ where $\omega$ is a complex number such that $\omega^3 = 1$. Hmm, let me start by recalling what the roots of the equation $\omega^3 = 1$ are.

First, I know that the solutions to $\omega^3 = 1$ are the cube roots of unity. There's one real root and two complex roots. The real root is obviously 1 because $1^3 = 1$. The other two roots are complex numbers, often denoted as $\omega$ and $\omega^2$, which are complex conjugates of each other.

So, the three cube roots of unity are 1, $\omega$, and $\omega^2$, where $\omega = e^{2\pi i /3}$ and $\omega^2 = e^{4\pi i /3}$. These satisfy the equation $1 + \omega + \omega^2 = 0$. That's a key identity I remember. Maybe that will come in handy later.

Now, the problem is asking for the value of $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me note that $\omega$ can be either of the complex roots or the real root 1. Wait, but if $\omega$ is 1, then the expression becomes $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. But the problem says "all possible values," so maybe there are different values depending on which root $\omega$ is? Wait, but the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be any of the three roots. However, if $\omega$ is 1, then the expression is 1. If $\omega$ is a complex root, then maybe the expression evaluates to something else?

Wait, but maybe the problem is considering $\omega$ as a non-real root? Because if $\omega$ is 1, then the expression is 1, but maybe the problem wants the values when $\omega$ is not 1? Wait, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$", so $\omega$ can be any of the three roots. Therefore, we need to check all three possibilities. However, maybe when $\omega$ is 1, the expression is 1, and when $\omega$ is the other roots, maybe the expression is the same? Or different?

Wait, let me check. Let me first compute the expression for $\omega = 1$. Then, as I said, it's 1. Now, let me compute it for $\omega$ being a complex root. Let me take $\omega$ as $e^{2\pi i /3}$, which is $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then $\omega^2$ is $e^{4\pi i /3}$, which is $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

So, let me compute $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me compute each term separately.

First, compute $1 + \omega$. Since $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, adding 1 gives $1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. Similarly, $1 + \omega^2 = 1 + (-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.

So, the first term is $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$ and the second term is $\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}}$. Let me compute these fractions.

To compute $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$, I can multiply numerator and denominator by the complex conjugate of the denominator. The complex conjugate of $\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is $\frac{1}{2} - i\frac{\sqrt{3}}{2}$. So:

$$
\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} \times \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{1} = \frac{1}{2} - i\frac{\sqrt{3}}{2}
$$

Similarly, the second term:

$$
\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} \times \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{1} = \frac{1}{2} + i\frac{\sqrt{3}}{2}
$$

Adding these two results together:

$$
\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \frac{1}{2} + (-i\frac{\sqrt{3}}{2} + i\frac{\sqrt{3}}{2}) = 1 + 0 = 1
$$

Wait, so even when $\omega$ is a complex root, the expression still evaluates to 1? But that's the same as when $\omega = 1$. So does that mean that regardless of which root $\omega$ is, the expression is always 1? But that seems strange. Let me check again.

Wait, if $\omega = 1$, then the expression is 1. If $\omega$ is a complex root, then the expression is also 1. Therefore, the only possible value is 1? But the problem says "Find all possible values," so maybe there's only one value? But maybe I made a mistake here?

Wait, let me try another approach. Let me use algebra instead of plugging in specific values. Let me consider that $\omega^3 = 1$, so $\omega \neq 1$ (if we consider the non-real roots), but maybe I can use the identity $1 + \omega + \omega^2 = 0$. Let me try to manipulate the expression.

Let me denote the expression as $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me combine the two fractions:

First, find a common denominator, which would be $(1 + \omega)(1 + \omega^2)$. So:

$$
E = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)} = \frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

Now, since $1 + \omega + \omega^2 = 0$, we know that $\omega + \omega^2 = -1$. Therefore, the numerator becomes $2 + (-1) = 1$. So the numerator is 1.

Now, let me compute the denominator: $(1 + \omega)(1 + \omega^2)$. Let me expand this:

$$
(1 + \omega)(1 + \omega^2) = 1 \cdot 1 + 1 \cdot \omega^2 + \omega \cdot 1 + \omega \cdot \omega^2 = 1 + \omega^2 + \omega + \omega^3
$$

But $\omega^3 = 1$, so substituting that in:

$$
1 + \omega^2 + \omega + 1 = 2 + \omega + \omega^2
$$

Again, since $\omega + \omega^2 = -1$, this becomes $2 + (-1) = 1$. Therefore, the denominator is 1.

Therefore, the entire expression $E = \frac{1}{1} = 1$.

Wait, so regardless of which root $\omega$ is, the expression simplifies to 1? But that seems to contradict my initial thought that maybe there are different values. But according to this algebraic manipulation, it's always 1.

But let me check again with $\omega = 1$. If $\omega = 1$, then the original expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So even in that case, it's 1. Therefore, regardless of which root $\omega$ is, the value is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I made a mistake here? Let me think again.

Alternatively, perhaps there's a different approach. Let me consider that $\omega$ is a primitive cube root of unity, so $\omega \neq 1$. But even so, the expression still evaluates to 1.

Alternatively, maybe there's a mistake in my algebra? Let me check again:

Starting with $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

Combine the fractions:

Numerator: $ (1 + \omega^2) + (1 + \omega) = 2 + \omega + \omega^2 $

Denominator: $ (1 + \omega)(1 + \omega^2) $

As before, numerator is 1, denominator is 1. Therefore, E = 1.

Alternatively, maybe I can use substitution. Let me note that since $\omega^3 = 1$, then $\omega^2 = \omega^{-1}$, because $\omega^3 = 1$ implies $\omega^2 = \omega^{-1}$. Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Wait, that's a much simpler way! Because $\frac{1}{1 + \omega^{-1}} = \frac{\omega}{\omega + 1}$, since multiplying numerator and denominator by $\omega$:

$$
\frac{1}{1 + \omega^{-1}} = \frac{\omega}{\omega(1) + \omega \cdot \omega^{-1}} = \frac{\omega}{\omega + 1}
$$

Therefore, adding $\frac{1}{1 + \omega} + \frac{\omega}{1 + \omega} = \frac{1 + \omega}{1 + \omega} = 1$.

So regardless of the value of $\omega$ (as long as $\omega \neq -1$, but since $\omega^3 = 1$, $\omega$ can't be -1 because (-1)^3 = -1 ≠ 1). Therefore, this shows that the expression is always 1.

Therefore, regardless of which root $\omega$ is (whether it's 1 or the complex roots), the value of the expression is always 1. Therefore, the only possible value is 1.

But let me check once more with $\omega = 1$. If $\omega = 1$, then $\omega^2 = 1$, so the expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So that works.

Therefore, the answer is just 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I need to check if there are different values when considering different roots? But according to all the methods, it's always 1.

Alternatively, maybe there's a mistake in assuming that $\omega$ can be 1? Wait, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be 1, which is a complex number (even though it's real). So, in that case, the expression is 1.

Alternatively, maybe the problem is expecting different values? Wait, but according to all calculations, it's always 1.

Wait, let me check with another approach. Let me suppose that $\omega$ is a root of $x^3 - 1 = 0$, so $x^3 - 1 = (x - 1)(x^2 + x + 1)$. Therefore, the roots are 1, $\omega$, $\omega^2$ where $\omega^2 + \omega + 1 = 0$.

So, if I take the expression $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me compute this expression for $\omega$ being a root of $x^2 + x + 1 = 0$.

Let me denote $\omega$ as a root of $x^2 + x + 1 = 0$, so $\omega^2 = -\omega - 1$.

Let me compute $1 + \omega$ and $1 + \omega^2$.

First, $1 + \omega$ is just $1 + \omega$.

Then, $1 + \omega^2 = 1 + (-\omega - 1) = -\omega$.

Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{ - \omega} = \frac{1}{1 + \omega} - \frac{1}{\omega}
$$

Now, let me compute this. Let me find a common denominator for these two terms. The common denominator would be $\omega(1 + \omega)$.

So:

$$
\frac{\omega}{\omega(1 + \omega)} - \frac{1 + \omega}{\omega(1 + \omega)} = \frac{\omega - (1 + \omega)}{\omega(1 + \omega)} = \frac{\omega - 1 - \omega}{\omega(1 + \omega)} = \frac{-1}{\omega(1 + \omega)}
$$

Now, let me compute $\omega(1 + \omega)$. Since $\omega^2 + \omega + 1 = 0$, then $\omega(1 + \omega) = \omega + \omega^2 = -1$. Therefore, the denominator is -1. Therefore, the entire expression becomes:

$$
\frac{-1}{-1} = 1
$$

So again, we get 1. Therefore, even when $\omega$ is a complex root, the expression is 1.

Therefore, regardless of which root $\omega$ is, the value of the expression is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I need to check if there are different values when considering different roots? But according to all calculations, it's always 1.

Alternatively, maybe there's a mistake in the problem statement? Or maybe I misread it? Let me check again.

The problem says: Let $\omega$ be a complex number such that $\omega^3 = 1$. Find all possible values of $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

So, since $\omega$ can be 1, $\omega$, or $\omega^2$, but in all cases, the expression evaluates to 1. Therefore, the only possible value is 1.

Alternatively, maybe the problem is expecting different values? Wait, but if $\omega$ is 1, then $\omega^2$ is also 1, so the expression is $\frac{1}{2} + \frac{1}{2} = 1$. If $\omega$ is a complex root, then $\omega^2$ is the other complex root, and as shown earlier, the expression is still 1. Therefore, regardless of which root $\omega$ is, the value is 1.

Therefore, the answer is just 1.

But maybe the problem is designed to have different values? Let me think again. Suppose that $\omega$ is a primitive cube root of unity, so $\omega \neq 1$. Then, the expression is 1. If $\omega = 1$, then the expression is also 1. Therefore, regardless, the value is 1.

Alternatively, maybe there's a different interpretation? For example, maybe the problem is considering $\omega$ as a non-real root? But even then, the expression still evaluates to 1.

Alternatively, maybe there's a mistake in my calculation? Let me check once more with numbers. Let me take $\omega = e^{2\pi i /3}$. Then, compute $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

Let me compute numerically. Let me compute $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then $1 + \omega = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. The reciprocal of this is $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$. Let me compute this numerically.

The modulus of $1 + \omega$ is $\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$. Therefore, the reciprocal has modulus 1. The angle of $1 + \omega$ is $\pi/3$ (since it's in the first quadrant, with real part 1/2 and imaginary part sqrt(3)/2). Therefore, the reciprocal has angle -π/3, so it's $\cos(-\pi/3) + i\sin(-\pi/3) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. Similarly, $1 + \omega^2$ is $\frac{1}{2} - i\frac{\sqrt{3}}{2}$, whose reciprocal is $\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Adding them together gives $\frac{1}{2} - i\frac{\sqrt{3}}{2} + \frac{1}{2} + i\frac{\sqrt{3}}{2} = 1$.

Therefore, numerically, it's 1.

Therefore, all methods confirm that the expression is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? Therefore, the answer is \boxed{1}.

But just to be thorough, let me check if there's a case where the expression could be different. Suppose someone thought that $\omega$ is only the non-real roots? But even then, as shown, the expression is still 1.

Alternatively, maybe there's a mistake in assuming that $\omega$ can be 1? But the problem says "complex number," and 1 is a complex number. So, it's allowed.

Alternatively, maybe the problem is in a context where $\omega$ is a primitive root? But even then, the expression still evaluates to 1.

Therefore, I think the answer is only 1.

**Final Answer**
\boxed{1}

To evaluate the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
where $\omega$ is a complex number such that $\omega^3 = 1$, we begin by analyzing the nature of the roots of the equation $\omega^3 = 1$.

---

### Step 1: Understanding the Roots of Unity

The equation $\omega^3 = 1$ has three roots, known as the **cube roots of unity**:

- $\omega_1 = 1$
- $\omega_2 = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
- $\omega_3 = e^{4\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

These roots satisfy the identity:
$$
1 + \omega + \omega^2 = 0
$$
This identity will be crucial in simplifying the expression.

---

### Step 2: Simplify the Expression

Let us denote the expression as:
$$
E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

We combine the two terms over a common denominator:
$$
E = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)} = \frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

From the identity $1 + \omega + \omega^2 = 0$, we know:
$$
\omega + \omega^2 = -1
$$
So the numerator becomes:
$$
2 + \omega + \omega^2 = 2 - 1 = 1
$$

Now compute the denominator:
$$
(1 + \omega)(1 + \omega^2) = 1 + \omega + \omega^2 + \omega^3
$$
Since $\omega^3 = 1$, this becomes:
$$
1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2 = 2 - 1 = 1
$$

Thus, the entire expression simplifies to:
$$
E = \frac{1}{1} = 1
$$

---

### Step 3: Verifying for All Roots

We have shown that the expression evaluates to 1 regardless of which root $\omega$ is chosen (whether it is 1, $\omega_2$, or $\omega_3$). This is because the identity $1 + \omega + \omega^2 = 0$ holds for all cube roots of unity, and the algebraic manipulation is valid for all such roots.

Even when $\omega = 1$, the expression becomes:
$$
\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1
$$

---

### Final Conclusion

No matter which cube root of unity $\omega$ is chosen, the value of the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
is always equal to 1.

---

### Final Answer

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find all possible values of the expression $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$ where $\omega$ is a complex number such that $\omega^3 = 1$. Hmm, let me start by recalling what the roots of the equation $\omega^3 = 1$ are.

First, I know that the solutions to $\omega^3 = 1$ are the cube roots of unity. There's one real root and two complex roots. The real root is obviously 1 because $1^3 = 1$. The other two roots are complex numbers, often denoted as $\omega$ and $\omega^2$, which are complex conjugates of each other.

So, the three cube roots of unity are 1, $\omega$, and $\omega^2$, where $\omega = e^{2\pi i /3}$ and $\omega^2 = e^{4\pi i /3}$. These satisfy the equation $1 + \omega + \omega^2 = 0$. That's a key identity I remember. Maybe that will come in handy later.

Now, the problem is asking for the value of $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me note that $\omega$ can be either of the complex roots or the real root 1. Wait, but if $\omega$ is 1, then the expression becomes $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. But the problem says "all possible values," so maybe there are different values depending on which root $\omega$ is? Wait, but the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be any of the three roots. However, if $\omega$ is 1, then the expression is 1. If $\omega$ is a complex root, then maybe the expression evaluates to something else?

Wait, but maybe the problem is considering $\omega$ as a non-real root? Because if $\omega$ is 1, then the expression is 1, but maybe the problem wants the values when $\omega$ is not 1? Wait, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$", so $\omega$ can be any of the three roots. Therefore, we need to check all three possibilities. However, maybe when $\omega$ is 1, the expression is 1, and when $\omega$ is the other roots, maybe the expression is the same? Or different?

Wait, let me check. Let me first compute the expression for $\omega = 1$. Then, as I said, it's 1. Now, let me compute it for $\omega$ being a complex root. Let me take $\omega$ as $e^{2\pi i /3}$, which is $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then $\omega^2$ is $e^{4\pi i /3}$, which is $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

So, let me compute $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me compute each term separately.

First, compute $1 + \omega$. Since $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, adding 1 gives $1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. Similarly, $1 + \omega^2 = 1 + (-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.

So, the first term is $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$ and the second term is $\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}}$. Let me compute these fractions.

To compute $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$, I can multiply numerator and denominator by the complex conjugate of the denominator. The complex conjugate of $\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is $\frac{1}{2} - i\frac{\sqrt{3}}{2}$. So:

$$
\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} \times \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{1} = \frac{1}{2} - i\frac{\sqrt{3}}{2}
$$

Similarly, the second term:

$$
\frac{1}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} \times \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{1} = \frac{1}{2} + i\frac{\sqrt{3}}{2}
$$

Adding these two results together:

$$
\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \frac{1}{2} + (-i\frac{\sqrt{3}}{2} + i\frac{\sqrt{3}}{2}) = 1 + 0 = 1
$$

Wait, so even when $\omega$ is a complex root, the expression still evaluates to 1? But that's the same as when $\omega = 1$. So does that mean that regardless of which root $\omega$ is, the expression is always 1? But that seems strange. Let me check again.

Wait, if $\omega = 1$, then the expression is 1. If $\omega$ is a complex root, then the expression is also 1. Therefore, the only possible value is 1? But the problem says "Find all possible values," so maybe there's only one value? But maybe I made a mistake here?

Wait, let me try another approach. Let me use algebra instead of plugging in specific values. Let me consider that $\omega^3 = 1$, so $\omega \neq 1$ (if we consider the non-real roots), but maybe I can use the identity $1 + \omega + \omega^2 = 0$. Let me try to manipulate the expression.

Let me denote the expression as $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me combine the two fractions:

First, find a common denominator, which would be $(1 + \omega)(1 + \omega^2)$. So:

$$
E = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)} = \frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

Now, since $1 + \omega + \omega^2 = 0$, we know that $\omega + \omega^2 = -1$. Therefore, the numerator becomes $2 + (-1) = 1$. So the numerator is 1.

Now, let me compute the denominator: $(1 + \omega)(1 + \omega^2)$. Let me expand this:

$$
(1 + \omega)(1 + \omega^2) = 1 \cdot 1 + 1 \cdot \omega^2 + \omega \cdot 1 + \omega \cdot \omega^2 = 1 + \omega^2 + \omega + \omega^3
$$

But $\omega^3 = 1$, so substituting that in:

$$
1 + \omega^2 + \omega + 1 = 2 + \omega + \omega^2
$$

Again, since $\omega + \omega^2 = -1$, this becomes $2 + (-1) = 1$. Therefore, the denominator is 1.

Therefore, the entire expression $E = \frac{1}{1} = 1$.

Wait, so regardless of which root $\omega$ is, the expression simplifies to 1? But that seems to contradict my initial thought that maybe there are different values. But according to this algebraic manipulation, it's always 1.

But let me check again with $\omega = 1$. If $\omega = 1$, then the original expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So even in that case, it's 1. Therefore, regardless of which root $\omega$ is, the value is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I made a mistake here? Let me think again.

Alternatively, perhaps there's a different approach. Let me consider that $\omega$ is a primitive cube root of unity, so $\omega \neq 1$. But even so, the expression still evaluates to 1.

Alternatively, maybe there's a mistake in my algebra? Let me check again:

Starting with $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

Combine the fractions:

Numerator: $ (1 + \omega^2) + (1 + \omega) = 2 + \omega + \omega^2 $

Denominator: $ (1 + \omega)(1 + \omega^2) $

As before, numerator is 1, denominator is 1. Therefore, E = 1.

Alternatively, maybe I can use substitution. Let me note that since $\omega^3 = 1$, then $\omega^2 = \omega^{-1}$, because $\omega^3 = 1$ implies $\omega^2 = \omega^{-1}$. Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Wait, that's a much simpler way! Because $\frac{1}{1 + \omega^{-1}} = \frac{\omega}{\omega + 1}$, since multiplying numerator and denominator by $\omega$:

$$
\frac{1}{1 + \omega^{-1}} = \frac{\omega}{\omega(1) + \omega \cdot \omega^{-1}} = \frac{\omega}{\omega + 1}
$$

Therefore, adding $\frac{1}{1 + \omega} + \frac{\omega}{1 + \omega} = \frac{1 + \omega}{1 + \omega} = 1$.

So regardless of the value of $\omega$ (as long as $\omega \neq -1$, but since $\omega^3 = 1$, $\omega$ can't be -1 because (-1)^3 = -1 ≠ 1). Therefore, this shows that the expression is always 1.

Therefore, regardless of which root $\omega$ is (whether it's 1 or the complex roots), the value of the expression is always 1. Therefore, the only possible value is 1.

But let me check once more with $\omega = 1$. If $\omega = 1$, then $\omega^2 = 1$, so the expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So that works.

Therefore, the answer is just 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I need to check if there are different values when considering different roots? But according to all the methods, it's always 1.

Alternatively, maybe there's a mistake in assuming that $\omega$ can be 1? Wait, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be 1, which is a complex number (even though it's real). So, in that case, the expression is 1.

Alternatively, maybe the problem is expecting different values? Wait, but according to all calculations, it's always 1.

Wait, let me check with another approach. Let me suppose that $\omega$ is a root of $x^3 - 1 = 0$, so $x^3 - 1 = (x - 1)(x^2 + x + 1)$. Therefore, the roots are 1, $\omega$, $\omega^2$ where $\omega^2 + \omega + 1 = 0$.

So, if I take the expression $E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$. Let me compute this expression for $\omega$ being a root of $x^2 + x + 1 = 0$.

Let me denote $\omega$ as a root of $x^2 + x + 1 = 0$, so $\omega^2 = -\omega - 1$.

Let me compute $1 + \omega$ and $1 + \omega^2$.

First, $1 + \omega$ is just $1 + \omega$.

Then, $1 + \omega^2 = 1 + (-\omega - 1) = -\omega$.

Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{ - \omega} = \frac{1}{1 + \omega} - \frac{1}{\omega}
$$

Now, let me compute this. Let me find a common denominator for these two terms. The common denominator would be $\omega(1 + \omega)$.

So:

$$
\frac{\omega}{\omega(1 + \omega)} - \frac{1 + \omega}{\omega(1 + \omega)} = \frac{\omega - (1 + \omega)}{\omega(1 + \omega)} = \frac{\omega - 1 - \omega}{\omega(1 + \omega)} = \frac{-1}{\omega(1 + \omega)}
$$

Now, let me compute $\omega(1 + \omega)$. Since $\omega^2 + \omega + 1 = 0$, then $\omega(1 + \omega) = \omega + \omega^2 = -1$. Therefore, the denominator is -1. Therefore, the entire expression becomes:

$$
\frac{-1}{-1} = 1
$$

So again, we get 1. Therefore, even when $\omega$ is a complex root, the expression is 1.

Therefore, regardless of which root $\omega$ is, the value of the expression is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? But maybe I need to check if there are different values when considering different roots? But according to all calculations, it's always 1.

Alternatively, maybe there's a mistake in the problem statement? Or maybe I misread it? Let me check again.

The problem says: Let $\omega$ be a complex number such that $\omega^3 = 1$. Find all possible values of $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

So, since $\omega$ can be 1, $\omega$, or $\omega^2$, but in all cases, the expression evaluates to 1. Therefore, the only possible value is 1.

Alternatively, maybe the problem is expecting different values? Wait, but if $\omega$ is 1, then $\omega^2$ is also 1, so the expression is $\frac{1}{2} + \frac{1}{2} = 1$. If $\omega$ is a complex root, then $\omega^2$ is the other complex root, and as shown earlier, the expression is still 1. Therefore, regardless of which root $\omega$ is, the value is 1.

Therefore, the answer is just 1.

But maybe the problem is designed to have different values? Let me think again. Suppose that $\omega$ is a primitive cube root of unity, so $\omega \neq 1$. Then, the expression is 1. If $\omega = 1$, then the expression is also 1. Therefore, regardless, the value is 1.

Alternatively, maybe there's a different interpretation? For example, maybe the problem is considering $\omega$ as a non-real root? But even then, the expression still evaluates to 1.

Alternatively, maybe there's a mistake in my calculation? Let me check once more with numbers. Let me take $\omega = e^{2\pi i /3}$. Then, compute $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$.

Let me compute numerically. Let me compute $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then $1 + \omega = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. The reciprocal of this is $\frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}$. Let me compute this numerically.

The modulus of $1 + \omega$ is $\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$. Therefore, the reciprocal has modulus 1. The angle of $1 + \omega$ is $\pi/3$ (since it's in the first quadrant, with real part 1/2 and imaginary part sqrt(3)/2). Therefore, the reciprocal has angle -π/3, so it's $\cos(-\pi/3) + i\sin(-\pi/3) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. Similarly, $1 + \omega^2$ is $\frac{1}{2} - i\frac{\sqrt{3}}{2}$, whose reciprocal is $\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Adding them together gives $\frac{1}{2} - i\frac{\sqrt{3}}{2} + \frac{1}{2} + i\frac{\sqrt{3}}{2} = 1$.

Therefore, numerically, it's 1.

Therefore, all methods confirm that the expression is always 1. Therefore, the only possible value is 1.

But the problem says "Find all possible values," so maybe there's only one value? Therefore, the answer is \boxed{1}.

But just to be thorough, let me check if there's a case where the expression could be different. Suppose someone thought that $\omega$ is only the non-real roots? But even then, as shown, the expression is still 1.

Alternatively, maybe there's a mistake in assuming that $\omega$ can be 1? But the problem says "complex number," and 1 is a complex number. So, it's allowed.

Alternatively, maybe the problem is in a context where $\omega$ is a primitive root? But even then, the expression still evaluates to 1.

Therefore, I think the answer is only 1.

**Final Answer**
\boxed{1}

To evaluate the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
where $\omega$ is a complex number such that $\omega^3 = 1$, we begin by analyzing the nature of the roots of the equation $\omega^3 = 1$.

---

### Step 1: Understanding the Roots of Unity

The equation $\omega^3 = 1$ has three roots, known as the **cube roots of unity**:

- $\omega_1 = 1$
- $\omega_2 = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
- $\omega_3 = e^{4\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

These roots satisfy the identity:
$$
1 + \omega + \omega^2 = 0
$$
This identity will be crucial in simplifying the expression.

---

### Step 2: Simplify the Expression

Let us denote the expression as:
$$
E = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

We combine the two terms over a common denominator:
$$
E = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)} = \frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

From the identity $1 + \omega + \omega^2 = 0$, we know:
$$
\omega + \omega^2 = -1
$$
So the numerator becomes:
$$
2 + \omega + \omega^2 = 2 - 1 = 1
$$

Now compute the denominator:
$$
(1 + \omega)(1 + \omega^2) = 1 + \omega + \omega^2 + \omega^3
$$
Since $\omega^3 = 1$, this becomes:
$$
1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2 = 2 - 1 = 1
$$

Thus, the entire expression simplifies to:
$$
E = \frac{1}{1} = 1
$$

---

### Step 3: Verifying for All Roots

We have shown that the expression evaluates to 1 regardless of which root $\omega$ is chosen (whether it is 1, $\omega_2$, or $\omega_3$). This is because the identity $1 + \omega + \omega^2 = 0$ holds for all cube roots of unity, and the algebraic manipulation is valid for all such roots.

Even when $\omega = 1$, the expression becomes:
$$
\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1
$$

---

### Final Conclusion

No matter which cube root of unity $\omega$ is chosen, the value of the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
is always equal to 1.

---

### Final Answer

$$
\boxed{1}
$$

Solution:
We can write
\begin{align*}
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} &= \frac{1 + \omega^2 + 1 + \omega}{(1 + \omega)(1 + \omega^2)} \\
&= \frac{2 + \omega + \omega^2}{1 + \omega + \omega^2 + \omega^3} \\
&= \frac{2 + \omega + \omega^2}{2 + \omega + \omega^2} \\
&= \boxed{1}.
\end{align*}