As a check on the formula  \(TraditionalForm
      \`\(B\_n\)(z)\  = \ 
        \(\((\(-1\))\)\^n\) \(\(Γ\_n\)(z)\)\) in line 17 of page 217, we first check a more general result about resultants.

A[p_] :=  Sum[a[i]*X^i,  {i, 0, p}]

B[q_]  :=  Sum[b[i]*X^i, {i, 0, q}]

res1 [p_, q_]  :=  Resultant[A[p], B[q], X];

reverseA[p_]  :=  Sum[a[i]*X^(p - i), {i, 0, p}]

reverseB[q_]  :=  Sum[b[i]*X^(q - i), {i, 0, q}]

res2[p_, q_]  :=  
        Resultant[reverseA[p], reverseB[q], X];

sign[p_, q_]  =  (-1)^(p*q)

iden[p_, q_]  :=  
      Expand[ res1[p, q]  -  sign[p, q]*res2[p, q] ]

iden[1, 1]

iden[1, 2]

iden[1, 3]

Table[ iden[p, q], {p, 1, 5}, {q, 1, 5}]

Here is the check for line 17 of page 217.

secondIdentity[n_]  :=  
      Expand[ res1[n, n]  - (-1)^n*res2[n, n] ]

Table[ secondIdentity[n], {n, 1, 5} ]