If `x=a + b, y= a alpha + b beta, z= a beta + b alpha`, where `alpha and beta` are complex cube roots of unity, then show that `xyz = a^(3) + b^(3)`

Answer

To solve the problem, we need to show that \( xyz = a^3 + b^3 \) given the definitions of \( x, y, z \) in terms of \( a, b \) and the complex cube roots of unity \( \alpha \) and \( \beta \).

### Step-by-step Solution:

1. **Define the Variables**:
We are given:
\[ x = a + b, \quad y = a\alpha + b\beta, \quad z = a\beta + b\alpha \]
where \( \alpha \) and \( \beta \) are the complex cube roots of unity.

2. **Identify the Roots of Unity**:
The complex cube roots of unity are:
\[ \alpha = \omega = e^{2\pi i / 3}, \quad \beta = \omega^2 = e^{-2\pi i / 3} \]
with the property that \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

3. **Calculate \( y \) and \( z \)**:
Substitute \( \alpha \) and \( \beta \) into \( y \) and \( z \):
\[ y = a\omega + b\omega^2, \quad z = a\omega^2 + b\omega \]

4. **Calculate the Product \( xyz \)**:
Now compute \( xyz \):
\[ xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega) \]

5. **Expand the Product**:
First, expand \( (a\omega + b\omega^2)(a\omega^2 + b\omega) \):
\[ (a\omega + b\omega^2)(a\omega^2 + b\omega) = a^2\omega\omega^2 + ab\omega + ab\omega^2 + b^2\omega^2\omega \]
Since \( \omega\omega^2 = \omega^3 = 1 \):
\[ = a^2 + ab(\omega + \omega^2) + b^2 \]
Using \( \omega + \omega^2 = -1 \):
\[ = a^2 - ab + b^2 \]

6. **Combine with \( x \)**:
Now, substitute back into \( xyz \):
\[ xyz = (a + b)(a^2 - ab + b^2) \]
Expanding this gives:
\[ = a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \]
\[ = a^3 - a^2b + ab^2 + b^3 - ab^2 + a^2b \]
Notice that the \( -a^2b \) and \( ab^2 \) terms cancel out:
\[ = a^3 + b^3 \]

7. **Conclusion**:
Thus, we have shown that:
\[ xyz = a^3 + b^3 \]
Hence proved.