Question:

Step 1: Establish the properties of the cube root of unity. Since $\omega$ is a non-real cube root of unity, we have: $$\omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0$$ These are fundamental properties that will be used throughout the solution. Step 2: Simplify the first fraction by recognizing a cyclic pattern. Let $p = a + b\omega + c\omega^2$. Multiply $p$ by $\omega^2$: $$\omega^2 p = a\omega^2 + b\omega^3 + c\omega^4$$ Since $\omega^3 = 1$, we have $\omega^4 = \omega$, so: $$\omega^2 p = a\omega^2 + b(1) + c\omega = b + c\omega + a\omega^2$$ This means the denominator of the first fraction equals $\omega^2 p$. Therefore: $$\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} = \frac{p}{\omega^2 p} = \frac{1}{\omega^2} = \omega$$ (using the fact that $\omega^{-2} = \omega$ since $\omega^3 = 1$) Step 3: Simplify the second fraction using the same cyclic pattern. Multiply $p$ by $\omega$: $$\omega p = a\omega + b\omega^2 + c\omega^3$$ Since $\omega^3 = 1$: $$\omega p = a\omega + b\omega^2 + c = c + a\omega + b\omega^2$$ This means the denominator of the second fraction equals $\omega p$. Therefore: $$\frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \frac{p}{\omega p} = \frac{1}{\omega} = \omega^2$$ (using the fact that $\omega^{-1} = \omega^2$ since $\omega^3 = 1$) Step 4: Add the two fractions to find the final answer. $$\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega + \omega^2$$ Using the fundamental property $1 + \omega + \omega^2 = 0$, we get: $$\omega + \omega^2 = -1$$ **Final Answer:** The value of the given expression is $\boxed{-1}$, which corresponds to **Option 4**.