`P(x)` is a polynomial such that `P(x)=f(x^3)+xg(x^3) , P(x)` is divided by `x^2+x+1` Find the value of `P(1)`.

Answer

To solve the problem, we need to find the value of \( P(1) \) given that \( P(x) = f(x^3) + x g(x^3) \) and that \( P(x) \) is divisible by \( x^2 + x + 1 \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
We are given \( P(x) = f(x^3) + x g(x^3) \). Since \( P(x) \) is divisible by \( x^2 + x + 1 \), the roots of this polynomial, which are \( \omega \) and \( \omega^2 \) (where \( \omega = e^{2\pi i / 3} \)), must satisfy \( P(\omega) = 0 \) and \( P(\omega^2) = 0 \).

2. **Finding \( P(\omega) \)**:
Substitute \( x = \omega \):
\[ P(\omega) = f(\omega^3) + \omega g(\omega^3) \]
Since \( \omega^3 = 1 \), we have:
\[ P(\omega) = f(1) + \omega g(1) = 0 \quad \text{(1)} \]

3. **Finding \( P(\omega^2) \)**:
Substitute \( x = \omega^2 \):
\[ P(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3) \]
Again, since \( (\omega^2)^3 = 1 \):
\[ P(\omega^2) = f(1) + \omega^2 g(1) = 0 \quad \text{(2)} \]

4. **Setting Up the System of Equations**:
From equations (1) and (2), we have:
\[ f(1) + \omega g(1) = 0 \quad \text{(1)} \]
\[ f(1) + \omega^2 g(1) = 0 \quad \text{(2)} \]

5. **Subtracting the Equations**:
Subtract equation (2) from equation (1):
\[ (\omega - \omega^2) g(1) = 0 \]
Since \( \omega - \omega^2 \neq 0 \), we conclude that:
\[ g(1) = 0 \]

6. **Substituting \( g(1) \) Back**:
Substitute \( g(1) = 0 \) back into either equation (1) or (2):
\[ f(1) + \omega \cdot 0 = 0 \implies f(1) = 0 \]

7. **Finding \( P(1) \)**:
Now we can find \( P(1) \):
\[ P(1) = f(1^3) + 1 \cdot g(1^3) = f(1) + g(1) = 0 + 0 = 0 \]

### Final Answer:
Thus, the value of \( P(1) \) is \( \boxed{0} \).