“…Let β ∈ [0, b -1 s ). Then the polynomial operator pencils are invertible on the imaginary axis and have the following representations:P j (λ; β; A) = F j (λ; β; A)F j (-λ; β; A), j = 1, n, where F j (λ; β; A) = n+1 s=1 λEω j,s (β)A ≡ n+1 m=0 α m,j (β)λ (n-m) A m ,Re ω j,s (β) < 0, s = 1, n + 1, and the numbers α m,j (β) > 0, m = 0, n + 1; α ν,j (β) > 0, ν = 0, n + 1, satisfy the following systems of equations:P j (λ; β; A) = (iλ) 2 E + A 2 n+1β(iλ) 2j A (2(n+1)-2j) j (β)α ν,j (β)λ (n-m+1) (-λ) (n-ν+1) A m+ν .Examples(i) if m = 1 and n = 2, then the following systems of equations are satisfied:(1) for j = 1, 2 (β) + α 2 1,2 (β) -3 = -β, 2α 1,2 (β)α2 2,2 (β) + 3 = 0.…”