Let `x = alpha + beta, y = alpha omega + beta omega ^(2), z = alpha omega ^(2) + beta omega, omega ` being an imaginary cube root of unity. Product of xyz is

Options

A. `alpha ^(2) + beta ^(2)`
B. `alpha ^(2) - beta ^(2)`
C. `alpha ^(3) + beta ^(3)`
D. `alpha ^(3) - beta ^(3)`

Answer

To solve the problem, we need to find the product \( xyz \) where:

- \( x = \alpha + \beta \)
- \( y = \alpha \omega + \beta \omega^2 \)
- \( z = \alpha \omega^2 + \beta \omega \)

Here, \( \omega \) is an imaginary cube root of unity, satisfying \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step-by-Step Solution:

1. **Write down the expressions for \( x \), \( y \), and \( z \)**:
\[ x = \alpha + \beta \]
\[ y = \alpha \omega + \beta \omega^2 \]
\[ z = \alpha \omega^2 + \beta \omega \]

2. **Calculate the product \( xyz \)**:
\[ xyz = (\alpha + \beta)(\alpha \omega + \beta \omega^2)(\alpha \omega^2 + \beta \omega) \]

3. **Expand the product**:
To expand \( (y)(z) \):
\[ yz = (\alpha \omega + \beta \omega^2)(\alpha \omega^2 + \beta \omega) \]
Expanding this gives:
\[ = \alpha^2 \omega^3 + \alpha \beta \omega^2 \cdot \omega + \beta \alpha \omega^2 \cdot \omega^2 + \beta^2 \omega^3 \]
Using \( \omega^3 = 1 \):
\[ = \alpha^2 + \alpha \beta \omega + \beta \alpha \omega^2 + \beta^2 \]

4. **Combine terms**:
\[ yz = \alpha^2 + \beta^2 + \alpha \beta (\omega + \omega^2) \]
Since \( \omega + \omega^2 = -1 \):
\[ = \alpha^2 + \beta^2 - \alpha \beta \]

5. **Now multiply by \( x \)**:
\[ xyz = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha \beta) \]
Expanding this:
\[ = \alpha(\alpha^2 + \beta^2 - \alpha \beta) + \beta(\alpha^2 + \beta^2 - \alpha \beta) \]
\[ = \alpha^3 + \alpha \beta^2 - \alpha^2 \beta + \beta \alpha^2 + \beta^3 - \alpha \beta^2 \]
\[ = \alpha^3 + \beta^3 \]

6. **Final Result**:
Therefore, the product \( xyz \) simplifies to:
\[ xyz = \alpha^3 + \beta^3 \]