I would like to find the Alexander polynomial of the link $L$, described below.
Let $K(q,r)$ be the $(q,r)$-torus knot embedded on a torus $V$. Inside the torus $V$, consider a smaller solid torus $V'$, that has the same longitude but a smaller meridian.
Then if $X=\mathbb{S}^3 \backslash K(q,r)$, the complement of the link $L$ is the same as $X \backslash V'$.
Take $a$ as the meridian of $V$ in $(\mathbb{S}^3\backslash V)\cup \partial V$, $b$ as the longitude of $V$ and $c$ as the meridian of $V'$. Since the generator $g$ of $\partial V$ is isotopic to $a^q$ and $b^rc^q$ in the two spaces respectively, using Seifert-van Kampen's Theorem we have: $$\pi_1(\mathbb{S}^3 \backslash L) = \langle a,b,c|a^{-q}b^rc^q\rangle$$
Let $s= a^{-q}b^rc^q$, then its Fox derivatives are: $$\begin{align*}\dfrac{\partial s}{\partial a} &= -a^{-q}\dfrac{1-a^q}{1-a}\\ \dfrac{\partial s}{\partial b} &= a^{-q}\dfrac{1-b^r}{1-b}\\ \dfrac{\partial s}{\partial c} &= a^{-q}b^r\dfrac{1-c^q}{1-c}\end{align*}$$
To abelianize the group, consider the linking numbers of the curves $a$, $b$ and $c$ with the link, so that $a = t^{r+1}$, $b =t^q$ and $c =t$.
The Alexander matrix would be 1 × 3 in order, so a generator of its 1st elementary ideal is the gcd of its three entries:
$$\gcd \left( \dfrac{1-t^{q(r+1)}}{1-t^{r+1}}, \, \dfrac{1-t^{qr}}{1-t^q}, \, \dfrac{1-t^q}{1-t}\right)$$
I have strong reasons to believe that the answer would be $\dfrac{(1-t)(1-t^{q(r+1)})}{1-t^{r+1}}$, but I have no idea how to get there.
Could someone please help? Thank you for your attention.