If `f(x)=(x^(4)+x^(2)+1)/(x^(2)-x+1)`, the value of `f(omega^(n))` (where `'omega'` is the non-real root of the equation `z^(3)=1` and 'n' is a multiple of 3), is ……….. .

Answer

To solve the problem, we need to evaluate the function \( f(x) = \frac{x^4 + x^2 + 1}{x^2 - x + 1} \) at \( x = \omega^n \), where \( \omega \) is a non-real root of the equation \( z^3 = 1 \) and \( n \) is a multiple of 3.

### Step-by-step Solution:

1. **Identify the roots of \( z^3 = 1 \)**:
The roots of the equation \( z^3 = 1 \) are \( 1, \omega, \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \). The non-real roots are \( \omega \) and \( \omega^2 \).

2. **Use properties of \( \omega \)**:
We know that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). Since \( n \) is a multiple of 3, we can express \( n \) as \( n = 3m \) for some integer \( m \). Therefore, \( \omega^n = \omega^{3m} = (\omega^3)^m = 1^m = 1 \).

3. **Evaluate \( f(\omega^n) \)**:
Since \( \omega^n = 1 \), we need to find \( f(1) \):
\[ f(1) = \frac{1^4 + 1^2 + 1}{1^2 - 1 + 1} \]

4. **Calculate \( f(1) \)**:
Substitute \( 1 \) into the function:
\[ f(1) = \frac{1 + 1 + 1}{1 - 1 + 1} = \frac{3}{1} = 3 \]

5. **Final result**:
Therefore, the value of \( f(\omega^n) \) is \( 3 \).

### Summary:
The value of \( f(\omega^n) \) where \( n \) is a multiple of 3 is \( 3 \).