Let `g(x)` and `h(x)` are two polynomials such that the polynomial P(x) `=g(x^(3))+xh(x^(3))` is divisible by `x^(2)+x+1`, then which one of the following is not true?

Options

A. `g(1)=h(1)=0`
B. g(1)=h(1) `ne 0`
C. `g(1) =-h(1)`
D. `g(1)+h(1)=0`

Answer

To solve the problem, we need to analyze the polynomial \( P(x) = g(x^3) + x h(x^3) \) under the condition that 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:
\[ x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2} \]
Let these roots be \( \omega = \frac{-1 + \sqrt{3}i}{2} \) and \( \omega^2 = \frac{-1 - \sqrt{3}i}{2} \).

2. **Evaluate \( P(\omega) \)**:
Since \( P(x) \) is divisible by \( x^2 + x + 1 \), we have:
\[ P(\omega) = g(\omega^3) + \omega h(\omega^3) = 0 \]
Note that \( \omega^3 = 1 \), thus:
\[ P(\omega) = g(1) + \omega h(1) = 0 \tag{1} \]

3. **Evaluate \( P(\omega^2) \)**:
Similarly, we evaluate \( P(\omega^2) \):
\[ P(\omega^2) = g((\omega^2)^3) + \omega^2 h((\omega^2)^3) = g(1) + \omega^2 h(1) = 0 \tag{2} \]

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

5. **Subtract the Two Equations**:
Subtract equation (2) from equation (1):
\[ (\omega - \omega^2) h(1) = 0 \]
Since \( \omega \neq \omega^2 \), we conclude:
\[ h(1) = 0 \tag{3} \]

6. **Substituting \( h(1) \) Back**:
Substitute \( h(1) = 0 \) back into equation (1):
\[ g(1) + \omega \cdot 0 = 0 \implies g(1) = 0 \tag{4} \]

7. **Conclusion**:
From equations (3) and (4), we find:
\[ g(1) = 0 \quad \text{and} \quad h(1) = 0 \]

### Final Result:
Both \( g(1) \) and \( h(1) \) are equal to zero. Therefore, the statement that is **not true** among the options provided is that \( g(1) \neq h(1) \).