The common roots of the equation `x^(3) + 2x^(2) + 2x + 1 = 0 and 1+ x^(2008)+ x^(2003) = 0` are (where `omega` is a complex cube root of unity)

Options

A. `omega, omega^(2)`
B. `1, omega^(2)`
C. `-1, -omega`
D. `omega, -omega^(2)`

Answer

To find the common roots of the equations \( x^3 + 2x^2 + 2x + 1 = 0 \) and \( 1 + x^{2008} + x^{2003} = 0 \), we will follow these steps:

### Step 1: Find the roots of the first equation
We start with the polynomial \( x^3 + 2x^2 + 2x + 1 \).

1. **Substituting \( x = 0 \)**:
\[ 0^3 + 2(0^2) + 2(0) + 1 = 1 \quad (\text{not a root}) \]

2. **Substituting \( x = 1 \)**:
\[ 1^3 + 2(1^2) + 2(1) + 1 = 1 + 2 + 2 + 1 = 6 \quad (\text{not a root}) \]

3. **Substituting \( x = -1 \)**:
\[ (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0 \quad (\text{is a root}) \]

Since \( x = -1 \) is a root, we can factor the polynomial as \( (x + 1)(Ax^2 + Bx + C) \).

### Step 2: Polynomial long division
We divide \( x^3 + 2x^2 + 2x + 1 \) by \( x + 1 \):

1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x + 1 \) to get \( x^3 + x^2 \).
3. Subtract:
\[ (2x^2 - x^2) + 2x + 1 = x^2 + 2x + 1 \]
4. Divide \( x^2 \) by \( x \) to get \( x \).
5. Multiply \( x \) by \( x + 1 \) to get \( x^2 + x \).
6. Subtract:
\[ (2x - x) + 1 = x + 1 \]
7. Divide \( x \) by \( x \) to get \( 1 \).
8. Multiply \( 1 \) by \( x + 1 \) to get \( x + 1 \).
9. Subtract:
\[ 0 \]

Thus, we have:
\[ x^3 + 2x^2 + 2x + 1 = (x + 1)(x^2 + x + 1) \]

### Step 3: Find the roots of \( x^2 + x + 1 = 0 \)
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} \]
This gives us:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]
These roots can be expressed as \( \omega \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \) (the complex cube roots of unity).

### Step 4: Check the second equation
Now we need to check the second equation \( 1 + x^{2008} + x^{2003} = 0 \).

1. Rewrite \( x^{2008} \) and \( x^{2003} \):
\[ x^{2008} = (x^3)^{669} \cdot x^1 = 1^{669} \cdot x = x \]
\[ x^{2003} = (x^3)^{667} \cdot x^2 = 1^{667} \cdot x^2 = x^2 \]

Thus, the equation simplifies to:
\[ 1 + x + x^2 = 0 \]

### Step 5: Verify roots
The equation \( 1 + x + x^2 = 0 \) has roots \( \omega \) and \( \omega^2 \) (the complex cube roots of unity).

### Conclusion
The common roots of the equations \( x^3 + 2x^2 + 2x + 1 = 0 \) and \( 1 + x^{2008} + x^{2003} = 0 \) are:
\[ \omega \text{ and } \omega^2 \]