How many different values can $ \angle ABC$ take, where $ A,B,C$ are distinct vertices of a cube?
To determine how many different values $\angle ABC$ can take, where $A, B, C$ are distinct vertices of a cube, we need to consider the geometric properties of the cube and the possible configurations of the vertices.

1. **Identify the possible configurations of vertices:**
   - **Vertices connected by an edge:** In this case, the triangle formed will have side lengths $1, 1, \sqrt{2}$.
   - **Vertices connected by a face diagonal:** In this case, the triangle formed will have side lengths $1, \sqrt{2}, \sqrt{3}$.
   - **Vertices connected by a space diagonal:** In this case, the triangle formed will have side lengths $\sqrt{2}, \sqrt{2}, \sqrt{2}$.

2. **Calculate the angles for each configuration:**
   - **Edge-connected vertices (1, 1, $\sqrt{2}$):**
     - The angle between the two edges of length 1 is $90^\circ$.
     - Using the Law of Cosines for the angle opposite the side $\sqrt{2}$:
       \[
       \cos(\theta) = \frac{1^2 + 1^2 - (\sqrt{2})^2}{2 \cdot 1 \cdot 1} = \frac{1 + 1 - 2}{2} = 0 \implies \theta = 90^\circ
       \]
     - The other two angles are $45^\circ$ each, since:
       \[
       \cos(\theta) = \frac{1^2 + (\sqrt{2})^2 - 1^2}{2 \cdot 1 \cdot \sqrt{2}} = \frac{1 + 2 - 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ
       \]

   - **Face diagonal-connected vertices (1, $\sqrt{2}$, $\sqrt{3}$):**
     - The angle between the edge of length 1 and the face diagonal $\sqrt{2}$ is $45^\circ$.
     - Using the Law of Cosines for the angle opposite the side $\sqrt{3}$:
       \[
       \cos(\theta) = \frac{1^2 + (\sqrt{2})^2 - (\sqrt{3})^2}{2 \cdot 1 \cdot \sqrt{2}} = \frac{1 + 2 - 3}{2\sqrt{2}} = 0 \implies \theta = 90^\circ
       \]
     - The other angle is $60^\circ$, since:
       \[
       \cos(\theta) = \frac{(\sqrt{2})^2 + (\sqrt{3})^2 - 1^2}{2 \cdot \sqrt{2} \cdot \sqrt{3}} = \frac{2 + 3 - 1}{2\sqrt{6}} = \frac{4}{2\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \implies \theta = 60^\circ
       \]

   - **Space diagonal-connected vertices ($\sqrt{2}, \sqrt{2}, \sqrt{2}$):**
     - All angles in this equilateral triangle are $60^\circ$.

3. **List all unique angles:**
   - From the above calculations, the unique angles are $45^\circ$, $60^\circ$, and $90^\circ$.

4. **Count the unique angles:**
   - There are 3 unique angles: $45^\circ$, $60^\circ$, and $90^\circ$.

The final answer is $\boxed{3}$.