The roots α, β of the equation 3x^2 + λx − 1 = 0, satisfy 1/α^2 + 1/β^2 = 15. The value of (α^3 + β^3)^2 is

Question

UPSC Quizroom · Accepted Answer

Sum and product of roots:
α + β = −λ / 3
αβ = −1 / 3

Use the given condition:
1/α² + 1/β² = (α² + β²) / (αβ)² = 15

α² + β² = (α + β)² − 2αβ = (−λ/3)² − 2(−1/3) = λ²/9 + 2/3
(αβ)² = (−1/3)² = 1/9

So:
1/α² + 1/β² = (λ²/9 + 2/3) / (1/9) = λ² + 6 = 15
⇒ λ² = 9 ⇒ λ = ±3

Compute α³ + β³:
α³ + β³ = (α + β)³ − 3αβ(α + β) = (−λ/3)³ − 3(−1/3)(−λ/3)
= −λ³/27 − λ/3 = −λ(λ² + 9)/27

For λ = 3 → α³ + β³ = −3(9 + 9)/27 = −2

For λ = −3 → α³ + β³ = 2

Square it:
(α³ + β³)² = 4

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