If `1, omega ` and `omega^(2)` are the three cube roots of unity, then what is the value of `((a w^(6) + b omega^(4) + c omega^(2)))/(( b + c omega^(10) + a omega^(8)))` ?

Options

A. A)`(a)/(b)`
B. B)b
C. C)`omega`
D. D)`omega^(2)`

Answer

To solve the problem, we need to simplify the expression given the properties of the cube roots of unity, which are \(1\), \(\omega\), and \(\omega^2\).

### Step-by-Step Solution:

1. **Understanding the Cube Roots of Unity**:
The cube roots of unity satisfy the equation \(x^3 = 1\). The roots are:
\[ 1, \quad \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3} \]
They have the properties:
\[ \omega^3 = 1, \quad \omega^2 + \omega + 1 = 0 \]

2. **Simplifying Powers of \(\omega\)**:
We can simplify the powers of \(\omega\) in the expression:
- \(\omega^6 = (\omega^3)^2 = 1^2 = 1\)
- \(\omega^4 = \omega^{3+1} = \omega^1 = \omega\)
- \(\omega^2\) remains \(\omega^2\)
- \(\omega^{10} = \omega^{9+1} = \omega^1 = \omega\)
- \(\omega^8 = \omega^{6+2} = \omega^2\)

3. **Substituting Back into the Expression**:
Now, we can substitute these simplifications back into the original expression:
\[ \frac{a \cdot \omega^6 + b \cdot \omega^4 + c \cdot \omega^2}{b + c \cdot \omega^{10} + a \cdot \omega^{8}} = \frac{a \cdot 1 + b \cdot \omega + c \cdot \omega^2}{b + c \cdot \omega + a \cdot \omega^2} \]
This simplifies to:
\[ \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} \]

4. **Identifying the Numerator and Denominator**:
We notice that both the numerator and denominator are the same:
- Numerator: \(a + b\omega + c\omega^2\)
- Denominator: \(b + c\omega + a\omega^2\)

5. **Final Simplification**:
Since the numerator and denominator are equal, the expression simplifies to:
\[ \frac{a + b\omega + c\omega^2}{a + b\omega + c\omega^2} = 1 \]

6. **Conclusion**:
Therefore, the value of the given expression is:
\[ \boxed{1} \]