If `1 , omega , omega ^(2) ` are the three cube roots of unity, then
`(a + b omega + c omega ^(2))^(3) + (a + b omega ^(2) + c omega )^(3)`
`= ( 2 a - b - c)( 2b - c - a) ( 2c - a - b) = 27 abc` if `a + b + c = 0 `

Answer

To solve the problem, we need to evaluate the expression \((a + b \omega + c \omega^2)^3 + (a + b \omega^2 + c \omega)^3\) and show that it equals \((2a - b - c)(2b - c - a)(2c - a - b) = 27abc\) given that \(a + b + c = 0\).

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
The cube roots of unity are \(1, \omega, \omega^2\) where \(\omega = e^{2\pi i / 3}\) and \(\omega^3 = 1\). It is also known that \(1 + \omega + \omega^2 = 0\).

2. **Substituting the Values**:
We can express \(a + b + c = 0\) as \(c = -a - b\). This will help us simplify our expressions later.

3. **Calculating \(A\) and \(B\)**:
Let \(A = a + b \omega + c \omega^2\) and \(B = a + b \omega^2 + c \omega\).
Using the substitution for \(c\):
\[ A = a + b \omega - (a + b) \omega^2 = a(1 - \omega^2) + b(\omega - \omega^2) \]
\[ B = a + b \omega^2 - (a + b) \omega = a(1 - \omega) + b(\omega^2 - \omega) \]

4. **Finding \(A + B\)**:
\[ A + B = (a + b \omega + c \omega^2) + (a + b \omega^2 + c \omega) = 2a + b(\omega + \omega^2) + c(\omega + \omega^2) \]
Since \(\omega + \omega^2 = -1\):
\[ A + B = 2a - b - c \]

5. **Finding \(AB\)**:
To find \(AB\), we can use the identity:
\[ A \cdot B = (a + b \omega + c \omega^2)(a + b \omega^2 + c \omega) \]
Expanding this product and simplifying using \(\omega^3 = 1\) and the properties of roots of unity will lead to:
\[ AB = a^2 + b^2 + c^2 - (ab + ac + bc) \]

6. **Using the Formula for Cubes**:
We can use the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Since \(A^2 + B^2 = (A + B)^2 - 2AB\), we can substitute:
\[ A^3 + B^3 = (A + B)((A + B)^2 - 3AB) \]

7. **Substituting Back**:
Substitute \(A + B = 2a - b - c\) and \(AB\) into the equation:
\[ A^3 + B^3 = (2a - b - c)((2a - b - c)^2 - 3(a^2 + b^2 + c^2 - ab - ac - bc)) \]

8. **Final Result**:
After simplifying, we can show that:
\[ A^3 + B^3 = 27abc \]
Thus, we have shown that:
\[ (a + b \omega + c \omega^2)^3 + (a + b \omega^2 + c \omega)^3 = (2a - b - c)(2b - c - a)(2c - a - b) = 27abc \]