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Orient the two components coherently through the twist region and assign meridian variables $x,y$. The link diagram is the torus link $T_{2,2n}$. It is obtained from the three-component torus link of part 2 by $-1/(n-1)$ rational Dehn surgery on the third component: removing its meridional disk adds $n-1$ full twists to the original single full twist. For $n=1$, this just means meridionally deleting the third component.

Write $z$ for the removed component's meridian. Its longitude is homologous to ${\mu }_1+{\mu }_2$, so the filling imposes $z=(xy{)}^{n-1}$. Under this substitution, the polynomial of part 2 becomes $(xy{)}^n-1$. The filling core is homologous, up to sign, to ${\mu }_1+{\mu }_2$. The Turaev-torsion Dehn-filling formula therefore removes the factor $xy-1$, givingFor positive $n$ this is the Laurent polynomial $1+xy+\cdots +(xy{)}^{n-1}$. For $n=1$ it is one, as for a Hopf link; for $n=0$ it is zero, as for the two-component unlink. For negative $n$ the displayed quotient is still a Laurent polynomial and agrees with the mirrored positive-twist answer up to a unit. These checks also fix the twist count: the exponent is $n$, rather than $n+1$.

The manifold $X_P$ is obtained from $X_{U,P}$ by filling the $U$-boundary along ${\mu }_U$. The filling core represents ${\iota }_{\ast }^{\partial X_U}[{\lambda }_U]$ in $H_1(X_P)$. The Turaev-torsion Dehn-filling formula therefore givesSince a knot exterior has first Betti number one, its normalized Turaev torsion isSubstitution proves