If the sum of the squares of the reciprocals of the roots α and β of
the equation 3x² + λx - 1 = 0 is 15, then 6(α³ + β³)² is equal to :

Solution

Maths sir · 4 steps

Rebuild everything from α + β and αβ; you only ever need λ².

1. Sum and product. For 3x² + λ x - 1 = 0, we have α + β = -λ/3 and αβ = -1/3, so (αβ)² = 1/9.

   1/α² + 1/β² = ((α+β)² - 2αβ)/(αβ)² = 9(λ²/9 + 2/3) = λ² + 6

2. Find lambda squared. Setting λ² + 6 = 15 gives λ² = 9. You do not need to know whether λ is 3 or -3.
3. Sum of cubes. α³ + β³ = (α+β)³ - 3αβ(α+β) = -λ³/27 - λ/3 = -λ(λ²/27 + 1/3). With λ² = 9 the bracket is 1/3 + 1/3 = 2/3.
4. So. (α³ + β³)² = λ² × 4/9 = 9 × 4/9 = 4. Then 6 × 4 = 24, which is option (b).

The idea to keep. Any symmetric expression in the roots can be rebuilt from α + β and αβ. If the final answer is squared, you only need λ², never λ itself.

Answer from the NTA final answer key. Solution by PrepWiser.