Prove that `a^3 + b^3 + c^3 – 3abc = (a + b + c) (a + bomega + comega^2) (a + bomega^2 + "c"omega)`, where `omega` is an imaginary cube root of unity.

Answer

To prove the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \), where \( \omega \) is a cube root of unity, we will start by expanding the right-hand side (RHS) and show that it simplifies to the left-hand side (LHS).

### Step 1: Understand the properties of \( \omega \)
Recall that \( \omega \) is a cube root of unity, which means:
1. \( \omega^3 = 1 \)
2. \( 1 + \omega + \omega^2 = 0 \)

### Step 2: Write down the RHS
The RHS is:
\[ (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \]

### Step 3: Expand the first two factors
First, we will expand \( (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \).

Using the distributive property:
\[ = a(a + b\omega^2 + c\omega) + b\omega(a + b\omega^2 + c\omega) + c\omega^2(a + b\omega^2 + c\omega) \]
Expanding each term:
\[ = a^2 + ab\omega^2 + ac\omega + ab\omega + b^2\omega^3 + bc\omega^2 + ac\omega^2 + bc\omega + c^2\omega^3 \]
Now, substituting \( \omega^3 = 1 \):
\[ = a^2 + ab\omega^2 + ac\omega + ab\omega + b^2 + bc\omega^2 + ac\omega^2 + bc\omega + c^2 \]

### Step 4: Combine like terms
Now, we combine the terms:
\[ = a^2 + b^2 + c^2 + (ab + ac)(\omega + \omega^2) + (bc + ab)(\omega + \omega^2) \]
Using \( \omega + \omega^2 = -1 \):
\[ = a^2 + b^2 + c^2 - (ab + ac + bc) \]

### Step 5: Multiply by \( (a + b + c) \)
Now, we multiply this result by \( (a + b + c) \):
\[ = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 6: Expand the product
Expanding this product:
\[ = a(a^2 + b^2 + c^2 - ab - ac - bc) + b(a^2 + b^2 + c^2 - ab - ac - bc) + c(a^2 + b^2 + c^2 - ab - ac - bc) \]
This will yield:
\[ = a^3 + b^3 + c^3 - 3abc \]

### Conclusion
Thus, we have shown that:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \]
This completes the proof.