<p>If \(\alpha\), \(\beta\) and \(\gamma\) are the roots of the equation \(x^3 + 3x^2 - 4x - 2 = 0\), then find the values of the following expressions:</p><p>(i) \(\alpha^2 + \beta^2 + \gamma^2\)</p><p>(ii) \(\alpha^3 + \beta^3 + \gamma^3\)</p><p>(iii) \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find sums of roots and their powers, then apply algebraic identities like (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα) and the identity α³+β³+γ³-3αβγ = (α+β+γ)(α²+β²+γ²-αβ-βγ-γα).
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>For x³ + 3x² - 4x - 2 = 0:</p><p>α + β + γ = -3</p><p>αβ + βγ + γα = -4</p><p>αβγ = 2</p><p><strong>Step 2(i): Find α² + β² + γ²</strong></p><p>Using (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)</p><p>(-3)² = α² + β² + γ² + 2(-4)</p><p>9 = α² + β² + γ² - 8</p><p>α² + β² + γ² = <strong>17</strong></p><p><strong>Step 2(ii): Find α³ + β³ + γ³</strong></p><p>Since α, β, γ are roots: α³ = -3α² + 4α + 2 (and similarly for β, γ)</p><p>Adding: α³ + β³ + γ³ = -3(α² + β² + γ²) + 4(α + β + γ) + 6</p><p>α³ + β³ + γ³ = -3(17) + 4(-3) + 6 = -51 - 12 + 6 = <strong>-9</strong></p><p><strong>Step 2(iii): Find 1/α + 1/β + 1/γ</strong></p><p>1/α + 1/β + 1/γ = (βγ + γα + αβ)/(αβγ) = (-4)/2 = <strong>-2</strong></p><p>∴ Answers: (i) 17, (ii) -9, (iii) 2</p>
Correct Answer: (i) 17, (ii) -9, (iii) 2