If ` alpha , beta ` are complex cube roots of unity and ` x = a + b, y = a alpha + b beta , z = a beta + b alpha `, then xyz =

Options

A. `(a + b^(3))`
B. ` a^(3) + b^(3)`
C. ` (a - b)^(3)`
D. ` a^(3) - b^(3)`

Answer

To solve the problem, we need to find the product \( xyz \) given the definitions of \( x, y, z \) in terms of complex cube roots of unity \( \alpha \) and \( \beta \).

### Step-by-Step Solution:

1. **Identify the complex cube roots of unity:**
The complex cube roots of unity are given by:
\[ \alpha = \omega, \quad \beta = \omega^2 \]
where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \). We know that:
\[ 1 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^3 = 1 \]

2. **Substitute the values of \( x, y, z \):**
Given:
\[ x = a + b, \quad y = a\alpha + b\beta = a\omega + b\omega^2, \quad z = a\beta + b\alpha = a\omega^2 + b\omega \]

3. **Calculate \( xyz \):**
We can express \( xyz \) as:
\[ xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega) \]

4. **Simplify \( (a\omega + b\omega^2)(a\omega^2 + b\omega) \):**
Let's expand this product:
\[ (a\omega + b\omega^2)(a\omega^2 + b\omega) = a^2\omega^3 + ab(\omega + \omega^2) + b^2\omega^3 \]
Since \( \omega^3 = 1 \):
\[ = a^2 + b^2 + ab(\omega + \omega^2) \]
Using \( \omega + \omega^2 = -1 \):
\[ = a^2 + b^2 - ab \]

5. **Combine everything:**
Now substituting back into \( xyz \):
\[ xyz = (a + b)(a^2 + b^2 - ab) \]

6. **Final expression:**
We can express \( a^2 + b^2 - ab \) in a different form:
\[ a^2 + b^2 - ab = \frac{(a+b)^2 - ab}{2} \]
Thus:
\[ xyz = (a + b)(a^2 + b^2 - ab) = (a + b)\left(\frac{(a+b)^2 - ab}{2}\right) \]

7. **Using the identity for cubes:**
We know that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Therefore:
\[ xyz = a^3 + b^3 \]

### Conclusion:
Thus, the final result is:
\[ xyz = a^3 + b^3 \]