If `f(x)=g(x^(3))+xh(x^(3))` is divisiblel by `x^(2)+x+1`, then

Options

A. g(x) is divisible by (x-1) but not h(x) but not h(x)
B. h(x) is divisible by (x-1) but not g(x)
C. both g(x) and h(x) are divisible by (x-1)
D. None of above

Answer

To solve the problem, we need to determine the conditions under which the function \( f(x) = g(x^3) + xh(x^3) \) 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:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2} \]
Let \( \omega = \frac{-1 + \sqrt{3}i}{2} \) and \( \omega^2 = \frac{-1 - \sqrt{3}i}{2} \).

2. **Evaluate \( f(\omega) \)**:
Since \( f(x) \) is divisible by \( x^2 + x + 1 \), it must equal zero at the roots. Therefore, we evaluate:
\[ f(\omega) = g(\omega^3) + \omega h(\omega^3) \]
Since \( \omega^3 = 1 \), we have:
\[ f(\omega) = g(1) + \omega h(1) = 0 \quad \text{(Equation 1)} \]

3. **Evaluate \( f(\omega^2) \)**:
Similarly, we evaluate \( f(\omega^2) \):
\[ f(\omega^2) = g((\omega^2)^3) + \omega^2 h((\omega^2)^3) \]
Again, since \( \omega^2 \) cubed is also 1:
\[ f(\omega^2) = g(1) + \omega^2 h(1) = 0 \quad \text{(Equation 2)} \]

4. **Set Up the Equations**:
From Equation 1:
\[ g(1) + \omega h(1) = 0 \quad \text{(1)} \]
From Equation 2:
\[ g(1) + \omega^2 h(1) = 0 \quad \text{(2)} \]

5. **Solve the System of Equations**:
We can subtract Equation 1 from Equation 2:
\[ (g(1) + \omega^2 h(1)) - (g(1) + \omega h(1)) = 0 \]
This simplifies to:
\[ (\omega^2 - \omega) h(1) = 0 \]
Since \( \omega \neq \omega^2 \), we conclude that:
\[ h(1) = 0 \]

6. **Substituting Back**:
Substitute \( h(1) = 0 \) back into Equation 1:
\[ g(1) + \omega \cdot 0 = 0 \Rightarrow g(1) = 0 \]

### Conclusion:
Thus, for \( f(x) \) to be divisible by \( x^2 + x + 1 \), both \( g(1) = 0 \) and \( h(1) = 0 \) must hold.