A group of $6$ friends sit in the back row of an otherwise empty movie theater. Each row in the theater contains $8$ seats. Euler and Gauss are best friends, so they must sit next to each other, with no empty seat between them. However, Lagrange called them names at lunch, so he cannot sit in an adjacent seat to either Euler or Gauss. In how many different ways can the $6$ friends be seated in the back row?

$\text{(A) }2520\qquad\text{(B) }3600\qquad\text{(C) }4080\qquad\text{(D) }5040\qquad\text{(E) }7200$
1. **Identify the constraints and total seats:**
   - There are 8 seats in the row.
   - Euler (E) and Gauss (G) must sit next to each other.
   - Lagrange (L) cannot sit next to either Euler or Gauss.
   - We need to find the number of ways to seat 6 friends under these constraints.

2. **Treat Euler and Gauss as a single unit:**
   - Since Euler and Gauss must sit next to each other, we can treat them as a single "block" or "unit".
   - This "block" can be in the form (E, G) or (G, E), giving us 2 arrangements for the block itself.

3. **Calculate the possible positions for the Euler-Gauss block:**
   - The block occupies 2 seats, so there are \(8 - 2 + 1 = 7\) possible positions for the block in the row.
   - However, we need to consider the positions where Lagrange cannot sit next to the block.

4. **Case 1: Euler and Gauss sit on the seats on either edge:**
   - If the block is at the edge, there are 2 possible positions: (1, 2) or (7, 8).
   - For each of these positions, the block can be (E, G) or (G, E), giving us \(2 \times 2 = 4\) arrangements.
   - The remaining 4 friends (including Lagrange) need to be seated in the remaining 6 seats.
   - Lagrange cannot sit in the seats adjacent to the block, so we have 4 choices for Lagrange.
   - The remaining 3 friends can be seated in the remaining 5 seats in \(5 \times 4 \times 3 = 60\) ways.
   - Total for this case: \(4 \times 4 \times 60 = 960\) ways.

5. **Case 2: Euler and Gauss don't sit on the edge:**
   - The block can be in positions (2, 3), (3, 4), (4, 5), (5, 6), or (6, 7), giving us 5 positions.
   - For each position, the block can be (E, G) or (G, E), giving us \(5 \times 2 = 10\) arrangements.
   - Lagrange cannot sit in the seats adjacent to the block, so we have 4 choices for Lagrange.
   - The remaining 3 friends can be seated in the remaining 5 seats in \(5 \times 4 \times 3 = 60\) ways.
   - Total for this case: \(10 \times 4 \times 60 = 2400\) ways.

6. **Sum the total number of ways:**
   - Total number of ways = \(960 + 2400 = 3360\).

The final answer is \(\boxed{3360}\)