<p>If <em>α</em>, <em>β</em>, <em>γ</em> are the roots of equation \(x^3 - 3x^2 + 3x + 7 = 0\) and <em>ω</em> is a cube root of unity, then find the value of \(\dfrac{\alpha - 1}{\beta - 1} + \dfrac{\beta - 1}{\gamma - 1} + \dfrac{\gamma - 1}{\alpha - 1}\).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find relationships between roots, then substitute y = x - 1 to transform the equation and find the new roots. The sum of ratios can be expressed using the coefficients of the transformed equation.
<p><strong>Step 1: Apply Vieta's formulas to the original equation</strong></p><p>Given: x³ - 3x² + 3x + 7 = 0 with roots α, β, γ</p><p>From Vieta's formulas:</p><p>• α + β + γ = 3</p><p>• αβ + βγ + γα = 3</p><p>• αβγ = -7</p><p><strong>Step 2: Transform the equation using y = x - 1</strong></p><p>Substitute x = y + 1 into the original equation:</p><p>(y+1)³ - 3(y+1)² + 3(y+1) + 7 = 0</p><p>Expanding:</p><p>y³ + 3y² + 3y + 1 - 3(y² + 2y + 1) + 3y + 3 + 7 = 0</p><p>y³ + 3y² + 3y + 1 - 3y² - 6y - 3 + 3y + 3 + 7 = 0</p><p>y³ + 8 = 0, or y³ = -8</p><p><strong>Step 3: Identify the new roots</strong></p><p>The roots of y³ + 8 = 0 are y³ = -8, so y = -2ω^k where ω is a cube root of unity and k = 0, 1, 2</p><p>Therefore: α - 1 = -2, β - 1 = -2ω, γ - 1 = -2ω²</p><p>(or any permutation of these values)</p><p><strong>Step 4: Compute the required sum</strong></p><p>Let a = α - 1 = -2, b = β - 1 = -2ω, c = γ - 1 = -2ω²</p><p>We need: a/b + b/c + c/a</p><p>= (-2)/(-2ω) + (-2ω)/(-2ω²) + (-2ω²)/(-2)</p><p>= 1/ω + ω/ω² + ω²/1</p><p>= 1/ω + 1/ω + ω²</p><p>= ω² + ω² + ω²</p><p>= 3ω²</p><p><strong>Step 5: Verify using ω³ = 1 and 1 + ω + ω² = 0</strong></p><p>Since 1/ω = ω² (as ω·ω² = ω³ = 1), the calculation is correct.</p><p>∴ Answer: 3ω²</p>
Correct Answer: 3ω²