The slope-$4/3$ pattern has winding number $4$, so the <Satellite formula for the Alexander polynomial> gives
$$
\Delta_{T_{3,2}^{(4,3)}}(t)
\doteq\Delta_{T_{3,2}}(t^4)\Delta_{T_{4,3}}(t).
$$
Using the stated <torus knot> formula in each factor,
$$
\boxed{
\Delta_{T_{3,2}^{(4,3)}}(t)
=
\frac{(1-t^4)(1-t^{24})}{(1-t^{12})(1-t^8)}
\frac{(1-t)(1-t^{12})}{(1-t^4)(1-t^3)}.}
$$
Up to a Laurent unit this may be simplified to $(1-t)(1-t^{24})/((1-t^8)(1-t^3))$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-141/4/d/solution