Let $\alpha=(1\ 6)(3\ 2\ 8)$ and $\beta=(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$
Present $\alpha^{-1}\beta$ as a product of disjoint cycles
$$\alpha^{-1} = (8\ 2\ 3)(1\ 6)$$
$$\alpha^{-1}\beta = (8\ 2\ 3)(1\ 6)(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$$
In Cauchy notation $$\alpha^{-1}\beta = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ 3 & 4 & 5 & 6 & 7 & 8 & 1 & 9 & 10& 2 \end{pmatrix}$$
I'm struggling taking this back to cycle notation, is my inverse wrong?
Edit: So I have worked on it myself, and corrected my mistake with Cauchy notation. I work right to left for cycles and have reached the answer
$$\alpha^{-1}\beta = (1\ 3\ 5\ 7)(2\ 4\ 6\ 8\ 9\ 10)$$
Now I have to find the order of this, using lcm(4,6) I got $$|\alpha^{-1}\beta|=12$$
We wish to calculate the following in disjoint cycle notation \begin{eqnarray*} \alpha^{-1} \beta = (8\ 2\ 3)(1\ 6)(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10). \end{eqnarray*} Start with the element $1$ and reading left to right, $1$ is mapped to $6$ in the second cycle & the $6$ is mapped to $3$ in the fourth cycle, so $1$ maps to $3$. (So far $(1\ 3\ \cdots$).
Next $3$, $3$ is mapped to $8$ in the first cyle, which is mapped to $9$ in the last cycle (So far $(1\ 3\ 9\ \cdots)$).
Next $9$ is mapped to $10$ (So far $(1\ 3\ 9\ 10\ \cdots)$).
Continuing gives \begin{eqnarray*} \alpha^{-1} \beta = (1\ 3\ 9\ 10\ 8\ 4)(2\ 5\ 7\ 6). \end{eqnarray*}