Find a formula for the Jones polynomial of the disjoint union of two diagrams $K_1$ and $K_2$

Question

Find a formula for the Jones polynomial of the disjoint union of two diagrams $K_1$ and $K_2$.

I am having a lot of trouble with this seemingly easy question. I understand that a Jones polynomial looks like this:

$$V_K(t) = L_K(t^{-1/4}), \text{where}$$ $$L_K(A) = (-A^3)^{- \omega (K)} \cdot \langle K \rangle$$

But I don't understand how to apply this to two disjoint diagrams. Any help would be appreciated.

Update: Online resources have been telling me to using the Skein relation, and so I feel as if the approach I'm taking is not good.

Answer 1

We have \begin{eqnarray} \langle K_1\cup K_2\rangle&=&\sum_{\sigma}\langle K_1\cup K_2|\sigma\rangle d^{\|\sigma\|}\\ &=& \sum_{\sigma_1,\sigma_2}\langle K_1\cup K_2|\sigma_1\cup\sigma_2\rangle d^{\|\sigma_1\cup\sigma_2\|}\\ &=& \sum_{\sigma_1,\sigma_2}\langle K_1|\sigma_1\rangle \langle K_2|\sigma_2\rangle d^{\|\sigma_1\|+\|\sigma_2\|+1}\\ &=&d \langle K_1\rangle\langle K_2\rangle. \end{eqnarray} Therefore $$V_{K_1\cup K_2}(t)=-(\sqrt{t}+1/\sqrt{t})V_{K_1}(t)V_{ K_2}(t).$$