Two regular tetrahedrons $A$ and $B$ are made with the 8 vertices of a unit cube. (this way is unique)

What's the volume of $A\cup B$?
1. **Labeling and Setup**:
   - Label the vertices of the unit cube as follows: \(A, B, C, D\) for the bottom face (counter-clockwise) and \(E, F, G, H\) for the top face (counter-clockwise).
   - Define the tetrahedrons \(T_1\) and \(T_2\) as \(T_1 \equiv ACFH\) and \(T_2 \equiv BDEG\).

2. **Centers of Cube Faces**:
   - Identify the centers of the cube faces: \(K, L, M, N, P, Q\) for the faces \(ABFE\) (front), \(BCGF\) (right), \(CDHG\) (back), \(DAEH\) (left), \(ABCD\) (bottom), and \(EFGH\) (top), respectively.

3. **Smaller Cube and Midpoints**:
   - Consider a smaller cube with body diagonal \(EO\) and side lengths \(\frac{a}{2}\). Let \(R, S, T\) be the midpoints of the edges \(AE, EF, HE\), respectively.
   - The vertices of the smaller cube are \(RKONESQT\).

4. **Common Volume of Tetrahedrons**:
   - The plane \(KNQ\) is the midplane of tetrahedron \(T_2\) and also a face of tetrahedron \(T_1\). Thus, the common volume inside the smaller cube is the tetrahedron \(KNQO\).
   - By considering 7 such smaller cubes, the common volume of \(T_1\) and \(T_2\) is the regular octahedron \(KLMNPQ\).

5. **Volume of the Octahedron**:
   - The volume of the octahedron \(KLMNPQ\) is given by:
     \[
     V(T_1 \cap T_2) = V(KLMNPQ) = \frac{PQ}{3} \cdot S(KLMN) = \frac{a}{3} \left(\frac{a \sqrt{2}}{2}\right)^2 = \frac{a^3}{6}
     \]

6. **Volume of a Regular Tetrahedron**:
   - The side length of the regular tetrahedron \(T_1\) is:
     \[
     d = AC = AB \sqrt{2} = a \sqrt{2}
     \]
   - The height \(h\) of the tetrahedron is:
     \[
     h = \sqrt{d^2 - \left(\frac{d \sqrt{3}}{3}\right)^2} = \sqrt{2a^2 - \frac{2a^2}{3}} = \frac{2a}{\sqrt{3}}
     \]
   - The volume \(V(T_1)\) is:
     \[
     V(T_1) = \frac{h}{3} S(ACF) = \frac{h}{3} \cdot \frac{d^2 \sqrt{3}}{4} = \frac{2a}{3 \sqrt{3}} \cdot \frac{2a^2 \sqrt{3}}{4} = \frac{a^3}{3}
     \]

7. **Volume of the Union of Tetrahedrons**:
   - The volume of the union \(V(T_1 \cup T_2)\) is:
     \[
     V(T_1 \cup T_2) = V(T_1) + V(T_2) - V(T_1 \cap T_2) = 2V(T_1) - V(T_1 \cap T_2) = 2 \cdot \frac{a^3}{3} - \frac{a^3}{6} = \frac{4a^3}{6} - \frac{a^3}{6} = \frac{3a^3}{6} = \frac{a^3}{2}
     \]
   - Since \(a = 1\), the volume is:
     \[
     V(T_1 \cup T_2) = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\).