What is the number of axes of rotational symmetry of a truncated cube and how would you work this out. Please help I have no idea and neither does anyone else I know.
Think of if you formed an axis and did a rotation, how would the cube not change.
I'll give you a hint: the axis has to go through the origin of the cube, and the points of interest are going to be such that you consider the diagonal pairs and the centre of the cube faces (where there are 6 cube faces and 2 pairs of diagonals).
Thinking about the axis of rotation going through the origin is the best place to start.
it seems to me the axes of rotational symmetry of a truncated cube (i am assuming you mean the 8 corners are truncated to form equilaterial triangular facets), are the same as the axes of rotational symmetries of the untruncated cube.
we have 3 axes going through the centers of opposing faces of the untruncated cube.
we have 4 axes going through the center of opposing facets (the triangle formed by the truncations).
we have 6 axes going through the midpoints diagonally opposite edges between truncations.