The roots α, β of the equation 3x² + λx − 1 = 0.

Sum of the roots: α + β = −λ/3  
Product of the roots: αβ = −1/3  

Given: 1/α² + 1/β² = 15  

=> (α² + β²) / (αβ)² = 15  

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

Substitute into the condition:  
(λ² + 6) / (9 × 1/9) = 15  
=> λ² + 6 = 15  
=> λ² = 9  
=> λ = +3 or −3  

Now,  
α³ + β³ = (α + β)³ − 3(αβ)(α + β)  
         = (−λ³ / 27) − 3(−1/3)(−λ/3)  
         = (−λ³ / 27) − (λ / 3)  
         = −(λ³ / 27 + λ / 3)  

Case 1: λ = +3  
α³ + β³ = −(27/27 + 3/3) = −(1 + 1) = −2  
(α³ + β³)² = (−2)² = 4  

Case 2: λ = −3  
α³ + β³ = −(−27/27 + (−3)/3) = −(−1 − 1) = 2  
(α³ + β³)² = (2)² = 4  

Final Answer:  
(α³ + β³)² = 4