h Numbering $8$ vertices of cube from $1$ to $8$ such that there is no 2 consecutive number adjacent How many ways are there to number $8$ vertices of a cube using numbers
from $1-8$ such that there are no two consecutive numbers adjacent on
the cube ($1$ and $8$ are considered to be consecutive).

I know that the answer is $480$ (with symmetries/etc), however, I am looking for a solution. I see that there are $\frac{8!}{8\times 3}=1680$ ways (without symmetry/etc) to number the cube. I can't use graph theory for the problem. It was originally a probability problem.
Without symmetries, there are only ten solutions.  If the cube is $\{0,1\}^3$, put 1 at $(0,0,0)$ and the numbers at $(1,0,0),(0,1,0)$ and $(0,0,1)$ in increasing order.  Don't forget reflection symmetry.
There is another symmetry by reversing the order of the digits, from $1,2,...8$ to $1,8,7,...,2$.  You might assume the number at $(1,1,1)$ is less than 6.  The number at $(1,1,1)$ can't be 3 as there would be nowhere to put 2.
Now plough through the options.  With the extra symmetry of the second paragraph, there are between five and ten solutions to find.
If the number at $(1,1,1)$ is 2, then the neighbours of $1$ are $3$ and two out of $4,5,6,7$.  If the number at $(1,1,1)$ is 4, then the neighbours of $1$ are $3,5$ and another one.  If the number at $(1,1,1)$ is 5, then the neighbours of $1$ are $4,6$ and either 3 or 7.  There aren't that many possibilities to check.
Without symmetries, there are only ten solutions.  If the cube is $\{0,1\}^3$, put 1 at $(0,0,0)$ and the numbers at $(1,0,0),(0,1,0)$ and $(0,0,1)$ in increasing order.  Don't forget reflection symmetry.
There is another symmetry by reversing the order of the digits, from $1,2,...8$ to $1,8,7,...,2$.  You might assume the number at $(1,1,1)$ is less than 6.  The number at $(1,1,1)$ can't be 3 as there would be nowhere to put 2.
Now plough through the options.  With the extra symmetry of the second paragraph, there are between five and ten solutions to find.
If the number at $(1,1,1)$ is 2, then the neighbours of $1$ are $3$ and two out of $4,5,6,7$.  If the number at $(1,1,1)$ is 4, then the neighbours of $1$ are $3,5$ and another one.  If the number at $(1,1,1)$ is 5, then the neighbours of $1$ are $4,6$ and either 3 or 7.  There aren't that many possibilities to check.