Let g(x) and h(x) be two polynomials with real coefficients. If `p(x) = g(x^(3)) + xh(x^(3))` is divisible by `x^(2) + x + 1`, then

Options

A. g(1) = 0, h(1) = 1
B. g(1) = 1, h(1) = 0
C. g(1) = 0, h(1) = 0
D. g(1) = 1, h(1) = 0

Answer

To solve the problem, we need to analyze the polynomial \( p(x) = g(x^3) + xh(x^3) \) and determine the conditions under which it is divisible by \( x^2 + x + 1 \).

### Step-by-Step Solution:

1. **Identify the Roots of the Divisor**:
The polynomial \( x^2 + x + 1 \) has roots which can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Let \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

2. **Evaluate \( p(\omega) \)**:
Since \( p(x) \) is divisible by \( x^2 + x + 1 \), we have:
\[ p(\omega) = g(\omega^3) + \omega h(\omega^3) = 0 \]
We know that \( \omega^3 = 1 \). Therefore:
\[ p(\omega) = g(1) + \omega h(1) = 0 \]
This gives us our first equation:
\[ g(1) + \omega h(1) = 0 \quad \text{(Equation 1)} \]

3. **Evaluate \( p(\omega^2) \)**:
Similarly, we evaluate \( p(\omega^2) \):
\[ p(\omega^2) = g((\omega^2)^3) + \omega^2 h((\omega^2)^3) = g(1) + \omega^2 h(1) = 0 \]
This gives us our second equation:
\[ g(1) + \omega^2 h(1) = 0 \quad \text{(Equation 2)} \]

4. **Set Up the System of Equations**:
Now we have two equations:
- From Equation 1: \( g(1) + \omega h(1) = 0 \)
- From Equation 2: \( g(1) + \omega^2 h(1) = 0 \)

5. **Subtract the Equations**:
Subtract Equation 2 from Equation 1:
\[ (g(1) + \omega h(1)) - (g(1) + \omega^2 h(1)) = 0 \]
This simplifies to:
\[ \omega h(1) - \omega^2 h(1) = 0 \implies h(1)(\omega - \omega^2) = 0 \]

6. **Analyze the Result**:
Since \( \omega \neq \omega^2 \), we conclude that:
\[ h(1) = 0 \]

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

### Final Result:
Thus, we have:
\[ g(1) = 0 \quad \text{and} \quad h(1) = 0 \]

### Conclusion:
The polynomials \( g(x) \) and \( h(x) \) must satisfy \( g(1) = 0 \) and \( h(1) = 0 \). Therefore, both \( g(1) \) and \( h(1) \) are equal to zero.