Let \[T_0=2, T_1=3, T_2=6,\] and for $n\ge 3$, \[T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}.\] The first few terms are \[2, 3, 6, 14, 40, 152, 784, 5158, 40576, 363392.\] Find a formula for $T_n$ of the form \[T_n=A_n+B_n,\] where $\{A_n\}$ and $\{B_n\}$ are well known sequences.
To find a formula for \( T_n \) of the form \( T_n = A_n + B_n \), where \( \{A_n\} \) and \( \{B_n\} \) are well-known sequences, we need to verify that \( A_n = n! \) and \( B_n = 2^n \) satisfy the given recurrence relation and initial conditions.

1. **Initial Conditions:**
   - \( T_0 = 2 \)
   - \( T_1 = 3 \)
   - \( T_2 = 6 \)

2. **Proposed Sequences:**
   - Let \( A_n = n! \)
   - Let \( B_n = 2^n \)

3. **Verify Initial Conditions:**
   - For \( n = 0 \):
     \[
     T_0 = A_0 + B_0 = 0! + 2^0 = 1 + 1 = 2
     \]
   - For \( n = 1 \):
     \[
     T_1 = A_1 + B_1 = 1! + 2^1 = 1 + 2 = 3
     \]
   - For \( n = 2 \):
     \[
     T_2 = A_2 + B_2 = 2! + 2^2 = 2 + 4 = 6
     \]

4. **Verify Recurrence Relation:**
   - Given recurrence relation:
     \[
     T_n = (n+4)T_{n-1} - 4nT_{n-2} + (4n-8)T_{n-3}
     \]
   - Substitute \( T_n = n! + 2^n \):
     \[
     T_n = (n! + 2^n)
     \]
   - Check if \( A_n = n! \) and \( B_n = 2^n \) satisfy the recurrence:
     \[
     (n! + 2^n) = (n+4)(n-1)! + 2^{n-1}) - 4n((n-2)! + 2^{n-2}) + (4n-8)((n-3)! + 2^{n-3})
     \]

5. **Simplify and Verify:**
   - For \( A_n = n! \):
     \[
     n! = (n+4)(n-1)! - 4n(n-2)! + (4n-8)(n-3)!
     \]
     This is a known identity for factorials.
   - For \( B_n = 2^n \):
     \[
     2^n = (n+4)2^{n-1} - 4n2^{n-2} + (4n-8)2^{n-3}
     \]
     This can be verified by expanding and simplifying the terms.

Since both \( A_n = n! \) and \( B_n = 2^n \) satisfy the initial conditions and the recurrence relation, we conclude that:

\[
T_n = n! + 2^n
\]

The final answer is \( \boxed{ T_n = n! + 2^n } \)