If `1, omega, omega^(2)` are three cube roots of unity, show that `(a + omega b+ omega^(2)c) (a + omega^(2)b+ omega c)= a^(2) + b^(2) + c^(2)- ab- bc - ca`

Answer

To solve the problem, we need to show that:

\[ (a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) = a^2 + b^2 + c^2 - ab - bc - ca \]

where \(1, \omega, \omega^2\) are the cube roots of unity, satisfying \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step 1: Expand the Left-Hand Side

We start by expanding the left-hand side:

\[ (a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) \]

Using the distributive property (FOIL method), we get:

\[ = a(a + \omega^2 b + \omega c) + \omega b(a + \omega^2 b + \omega c) + \omega^2 c(a + \omega^2 b + \omega c) \]

Calculating each term:

1. \( a(a + \omega^2 b + \omega c) = a^2 + a\omega^2 b + a\omega c \)
2. \( \omega b(a + \omega^2 b + \omega c) = \omega ab + \omega^3 b^2 + \omega^2 bc = \omega ab + b^2 + \omega^2 bc \) (since \( \omega^3 = 1 \))
3. \( \omega^2 c(a + \omega^2 b + \omega c) = \omega^2 ac + \omega^2 b^2 + \omega^3 c^2 = \omega^2 ac + \omega^2 b^2 + c^2 \)

Now, combining all these terms, we have:

\[ = a^2 + (a\omega^2 b + \omega ab) + (a\omega c + \omega^2 ac) + (b^2 + \omega^2 b^2) + (c^2 + \omega^2 bc) \]

### Step 2: Combine Like Terms

Now we can combine the terms:

\[ = a^2 + b^2 + c^2 + (a\omega^2 b + \omega ab) + (a\omega c + \omega^2 ac) + (\omega^2 bc + b^2) \]

### Step 3: Simplify Using Properties of \(\omega\)

Using the properties of \(\omega\):

- \( \omega + \omega^2 = -1 \)
- \( \omega^2 + \omega = -1 \)

We can simplify the combinations:

\[ = a^2 + b^2 + c^2 + (ab(\omega + \omega^2)) + (ac(\omega + \omega^2)) + (bc(\omega + \omega^2)) \]

Substituting \( \omega + \omega^2 = -1 \):

\[ = a^2 + b^2 + c^2 - ab - ac - bc \]

### Conclusion

Thus, we have shown that:

\[ (a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) = a^2 + b^2 + c^2 - ab - ac - bc \]

Hence, the statement is proved.