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Theory of Equations: Roots Equation 111x 11x Equals

JEE Maths question with a full step-by-step solution.

Question
If α,β,γ\alpha, \beta, \gamma are roots of the equation 111x3−11x−1=0111x^3 - 11x - 1 = 0, then (αβ)−2+(βγ)−2+(γα)−2(\alpha\beta)^{-2} + (\beta\gamma)^{-2} + (\gamma\alpha)^{-2} equals
A23322332
B13311331
C12101210
D24422442correct
Solution
Step 1: By Vieta's: α+β+γ=0\alpha+\beta+\gamma = 0, αβ+βγ+γα=−11111\alpha\beta+\beta\gamma+\gamma\alpha = -\dfrac{11}{111}, αβγ=1111\alpha\beta\gamma = \dfrac{1}{111}. Step 2:
(αβ)−2+(βγ)−2+(γα)−2=γ2+α2+β2(αβγ)2(\alpha\beta)^{-2}+(\beta\gamma)^{-2}+(\gamma\alpha)^{-2} = \frac{\gamma^2+\alpha^2+\beta^2}{(\alpha\beta\gamma)^2}
 Step 3:
α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)=0−2(−11111)=22111\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 - 2\left(-\frac{11}{111}\right) = \frac{22}{111}
 Step 4:
22/111(1/111)2=22111⋅1112=22×111=2442\frac{22/111}{(1/111)^2} = \frac{22}{111} \cdot 111^2 = 22 \times 111 = 2442
 Correct answer: (4)
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