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Theory of Equations: Roots Equation Satisfying

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

Question
If α,β,γ\alpha, \beta, \gamma be the roots of the equation x3+ax+a=0x^3 + ax + a = 0 (where a∈Ra \in \mathbb{R}, a≠0a \neq 0) satisfying α2β+β2γ+γ2α=−8\dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\gamma} + \dfrac{\gamma^2}{\alpha} = -8, then aa is
A−8-8correct
B−2-2
C22
D88
Solution
Step 1: By Vieta's: α+β+γ=0\alpha+\beta+\gamma = 0, αβ+βγ+γα=a\alpha\beta+\beta\gamma+\gamma\alpha = a, αβγ=−a\alpha\beta\gamma = -a. Step 2: Multiply numerator and denominator by αβγ\alpha\beta\gamma:
α2β+β2γ+γ2α=α3γ+αβ3+βγ3αβγ\frac{\alpha^2}{\beta}+\frac{\beta^2}{\gamma}+\frac{\gamma^2}{\alpha} = \frac{\alpha^3\gamma + \alpha\beta^3 + \beta\gamma^3}{\alpha\beta\gamma}
 Step 3: Use α3=−aα−a\alpha^3 = -a\alpha - a (from the equation) to evaluate the numerator:
α3γ+αβ3+βγ3=γ(−aα−a)+α(−aβ−a)+β(−aγ−a)\alpha^3\gamma + \alpha\beta^3 + \beta\gamma^3 = \gamma(-a\alpha-a) + \alpha(-a\beta-a) + \beta(-a\gamma-a)
=−a(αγ+αβ+βγ)−a(γ+α+β)=−a⋅a−a⋅0=−a2= -a(\alpha\gamma+\alpha\beta+\beta\gamma) - a(\gamma+\alpha+\beta) = -a \cdot a - a \cdot 0 = -a^2
 Step 4:
α2β+β2γ+γ2α=−a2−a=a\frac{\alpha^2}{\beta}+\frac{\beta^2}{\gamma}+\frac{\gamma^2}{\alpha} = \frac{-a^2}{-a} = a
 Setting this equal to −8-8: a=−8a = -8. Correct answer: (1)
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