A tetrahedron $ABCD$ satisfies the following conditions: the edges $AB,AC$ and $AD$ are pairwise orthogonal, $AB=3$ and $CD=\sqrt2$. Find the minimum possible value of
$$BC^6+BD^6-AC^6-AD^6.$$
1. Denote \( x, y, z \) as the lengths \( AB, AC, AD \) respectively. We are given that:
   \[
   x = 3
   \]
   and since \( AB, AC, \) and \( AD \) are pairwise orthogonal, we can use the Pythagorean theorem in three dimensions to find:
   \[
   y^2 + z^2 = CD^2 = 2
   \]

2. We want to find the minimum value of:
   \[
   BC^6 + BD^6 - AC^6 - AD^6
   \]
   Since \( AB, AC, \) and \( AD \) are orthogonal, we can use the coordinates \( A = (0,0,0) \), \( B = (3,0,0) \), \( C = (0,y,0) \), and \( D = (0,0,z) \).

3. Calculate the distances \( BC \) and \( BD \):
   \[
   BC = \sqrt{(3-0)^2 + (0-y)^2 + (0-0)^2} = \sqrt{9 + y^2}
   \]
   \[
   BD = \sqrt{(3-0)^2 + (0-0)^2 + (0-z)^2} = \sqrt{9 + z^2}
   \]

4. We need to find:
   \[
   BC^6 + BD^6 - AC^6 - AD^6
   \]
   Substituting the values:
   \[
   BC^6 = (9 + y^2)^3
   \]
   \[
   BD^6 = (9 + z^2)^3
   \]
   \[
   AC^6 = y^6
   \]
   \[
   AD^6 = z^6
   \]

5. Therefore, the expression becomes:
   \[
   (9 + y^2)^3 + (9 + z^2)^3 - y^6 - z^6
   \]

6. Simplify the expression:
   \[
   (9 + y^2)^3 + (9 + z^2)^3 - y^6 - z^6 = 2 \cdot 9^3 + 3 \cdot 9^2 (y^2 + z^2) + 3 \cdot 9 (y^4 + z^4) + (y^6 + z^6) - y^6 - z^6
   \]
   \[
   = 2 \cdot 729 + 3 \cdot 81 \cdot 2 + 3 \cdot 9 (y^4 + z^4)
   \]
   \[
   = 1458 + 486 + 27 (y^4 + z^4)
   \]
   \[
   = 1944 + 27 (y^4 + z^4)
   \]

7. By the QM-AM inequality, we have:
   \[
   y^4 + z^4 \geq \frac{(y^2 + z^2)^2}{2} = \frac{2^2}{2} = 2
   \]
   Equality holds when \( y = z = 1 \).

8. Therefore, the minimum value is:
   \[
   1944 + 27 \cdot 2 = 1998
   \]

The final answer is \(\boxed{1998}\).