<p>If \(\omega\) is a non-real cube root of unity, then the value of \(\dfrac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \dfrac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the properties of cube roots of unity: ω³ = 1, 1 + ω + ω² = 0, and ω² = ω̄. Recognize that the numerator and denominators are cyclic permutations, allowing strategic algebraic manipulation using the constraint equation.
<p><strong>Step 1:</strong> Recall that for non-real cube root of unity ω: ω³ = 1, 1 + ω + ω² = 0, and ω² = ω̄ (conjugate).</p><p><strong>Step 2:</strong> Let P = a + bω + cω² (numerator of both terms). The denominators are cyclic permutations: D₁ = b + cω + aω² and D₂ = c + aω + bω².</p><p><strong>Step 3:</strong> Note that D₁ = ωP and D₂ = ω²P. This is because:<br>ωP = ω(a + bω + cω²) = aω + bω² + cω³ = aω + bω² + c = c + aω + bω² (reordering)<br>ω²P = ω²(a + bω + cω²) = aω² + bω³ + cω⁴ = aω² + b + cω = b + cω + aω² (reordering)</p><p><strong>Step 4:</strong> Substitute into the original expression:<br>P/(ωP) + P/(ω²P) = 1/ω + 1/ω² = ω²/ω³ + ω/ω³ = ω² + ω</p><p><strong>Step 5:</strong> Using 1 + ω + ω² = 0, we get ω + ω² = -1</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: A