24 ( ∑ i = 1 k α i λ i 4 ) − 24 ( ∑ i = 1 k α i λ i 3 ) ( ∑ i = 1 k α i λ i ) + 3 ( 2 ( 2 ( ∑ i = 1 k α i λ i 2 ) − ( ∑ i = 1 k α i λ i ) 2 ) + ( ∑ i = 1 k α i λ i ) 2 ) ( ∑ i = 1 k α i λ i ) 2 ( 2 ( ∑ i = 1 k α i λ i 2 ) − ( ∑ i = 1 k α i λ i ) 2 ) 2