Substituting the factored expression into the denominator, we see the common factor (1 -) in both the numerator and denominator. Since ≠ 1, we can cancel (1 -) to obtain (1)/(1 +).

When represents a complex cube root of unity, it satisfies the fundamental relation 1 + + ² = 0. From this identity, we can express 1 + as - ².

Substituting 1 + = - ² gives -(1)/( ²). Multiplying the numerator and denominator by, and using the fact that ³ = 1, we get -()/( ³) = -.

Simplify $\frac{1-\omega}{1-\omega^2}$ .

Answer: $\frac{1}{1+\omega}$ (or $-\omega$ if $\omega$ is a complex cube root of unity)

Step-by-step solution

Step 1: Factor the denominator

We recognize that the denominator $1 - \omega^2$ is a difference of two squares. Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$ with $a = 1$ and $b = \omega$ , we can factor it into $(1 - \omega)(1 + \omega)$ .

Step 2: Cancel common factors

Substituting the factored expression into the denominator, we see the common factor $(1 - \omega)$ in both the numerator and denominator. Since $\omega \neq 1$ , we can cancel $(1 - \omega)$ to obtain $\frac{1}{1 + \omega}$ .

Step 3: Apply cube root of unity properties

When $\omega$ represents a complex cube root of unity, it satisfies the fundamental relation $1 + \omega + \omega^2 = 0$ . From this identity, we can express $1 + \omega$ as $-\omega^2$ .

Step 4: Simplify using $\omega^3 = 1$

Substituting $1 + \omega = -\omega^2$ gives $-\frac{1}{\omega^2}$ . Multiplying the numerator and denominator by $\omega$ , and using the fact that $\omega^3 = 1$ , we get $-\frac{\omega}{\omega^3} = -\omega$ .

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Simplify 1−ω1−ω2\frac{1-\omega}{1-\omega^2}1−ω21−ω​.