Question

If $\omega$ is a cube root of unity, then $(1 + \omega - \omega^{2})(1 - \omega + \omega^{2})$ is

• A. 1
• B. 0
• C. 2
• D. 4

Hint:

Key properties of cube roots of unity: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Rewrite expressions using the second identity to simplify quickly.

Answer 1

The Correct Option is D

To solve the problem, we first need to understand the properties of cube roots of unity.

The cube roots of unity are solutions to the equation:

\(x^3 = 1\)

Apart from \(1\), the other two roots are denoted as \(\omega\) and \(\omega^{2}\), which satisfy the following properties:

• \(\omega^3 = 1\)
• \(\omega \neq 1\); \(\omega \neq \omega^{2}\)
• \(1 + \omega + \omega^{2} = 0\)

Now, we are given the expression \((1 + \omega - \omega^{2})(1 - \omega + \omega^{2})\). We will expand this using algebraic identities.

Let's set \(A = 1 + \omega - \omega^{2}\) and \(B = 1 - \omega + \omega^{2}\). We are to find \(A \times B = (1 + \omega - \omega^{2})(1 - \omega + \omega^{2})\).

Now use the distributive property (also known as the FOIL method for binomials):

\(AB = (1 \cdot 1) + (1 \cdot \omega^{2}) + (1 \cdot -\omega) + (\omega \cdot 1) + (\omega \cdot \omega^{2}) + (\omega \cdot -\omega) + (-\omega^{2} \cdot 1) + (-\omega^{2} \cdot \omega^{2}) + (-\omega^{2} \cdot -\omega)\)

Simplify:

• \(1 \cdot 1 = 1\)
• \(1 \cdot \omega^{2} = \omega^{2}\)
• \(1 \cdot -\omega = -\omega\)
• \(\omega \cdot 1 = \omega\)
• \(\omega \cdot \omega^{2} = \omega^{3} = 1\) (since \(\omega^3 = 1\))
• \(\omega \cdot -\omega = -\omega^{2}\)
• \(-\omega^{2} \cdot 1 = -\omega^{2}\)
• \(-\omega^{2} \cdot \omega^{2} = -(\omega^{4}) = -\omega\) (since \(\omega^{3} = 1\), \(\omega^{4} = \omega\))
• \(-\omega^{2} \cdot -\omega = \omega^{3} = 1\) (since \(\omega^3 = 1\))

Add up all terms: \(1 + \omega^{2} - \omega + \omega + 1 - \omega^{2} - \omega + 1 + 1\)

Combine like terms:

\(1 + 1 + 1 + 1 = 4\)

Thus, the expression evaluates to \(4\).

Therefore, the correct answer is Option 4: 4.