= Solution
Insert a kink on any component of an oriented <link diagram> for $L$. Switching the kink's crossing gives another <link diagram> of $L$; its <oriented smoothing> separates off a small <unknot> in a ball disjoint from the rest of $L$. The <skein relation> therefore gives
$$
(q-q^{-1})V(L)=(q^{1/2}-q^{-1/2})V(L\amalg U_1).
$$
Cancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the <Laurent polynomial> ring to obtain
$$
\boxed{V(L\amalg U_1)=(q^{1/2}+q^{-1/2})V(L)=V(U_2)V(L).}
$$
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