Given a cube of unit side. Let $A$ and $B$ be two opposite vertex. Determine the radius of the sphere, with center inside the cube, tangent to the three faces of the cube with common point $A$ and tangent to the three sides with common point $B$.
1. **Set up the coordinate system:**
   Place the cube in a coordinate system such that vertex \( A \) is at the origin \((0,0,0)\) and vertex \( B \) is at \((1,1,1)\).

2. **Equation of the sphere:**
   Let the center of the sphere be at \((r, r, r)\) and the radius of the sphere be \( r \). The equation of the sphere is:
   \[
   (x - r)^2 + (y - r)^2 + (z - r)^2 = r^2
   \]

3. **Tangency conditions:**
   The sphere is tangent to the three faces of the cube that intersect at \( A \). These faces are the planes \( x = 0 \), \( y = 0 \), and \( z = 0 \). For the sphere to be tangent to these planes, the distance from the center of the sphere to each plane must be equal to the radius \( r \). Therefore:
   \[
   r = r
   \]
   This condition is trivially satisfied.

4. **Tangency to the edges at \( B \):**
   The sphere is also tangent to the three edges of the cube that intersect at \( B \). These edges are along the lines \( x = 1 \), \( y = 1 \), and \( z = 1 \). For the sphere to be tangent to these lines, the distance from the center of the sphere to each line must be equal to the radius \( r \). The distance from the center \((r, r, r)\) to the line \( x = 1 \) is:
   \[
   \sqrt{(1 - r)^2 + (r - r)^2 + (r - r)^2} = |1 - r|
   \]
   Similarly, the distances to the lines \( y = 1 \) and \( z = 1 \) are also \( |1 - r| \).

5. **Equating the distances:**
   For the sphere to be tangent to all three lines, we must have:
   \[
   |1 - r| = r
   \]
   Solving this equation:
   \[
   1 - r = r \quad \text{or} \quad 1 - r = -r
   \]
   \[
   1 = 2r \quad \text{or} \quad 1 = 0 \quad (\text{which is not possible})
   \]
   Therefore:
   \[
   r = \frac{1}{2}
   \]

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