`49^(3)-30^(3)+(.......)^(3)=3 xx 49 xx 30 xx 19`

Answer

To solve the equation \(49^3 - 30^3 + (.......)^3 = 3 \times 49 \times 30 \times 19\), we can use the identity for the sum of cubes. The identity states that if \(a + b + c = 0\), then:

\[ a^3 + b^3 + c^3 = 3abc \]

### Step-by-Step Solution:

1. **Identify the terms**: We have \(a = 49\), \(b = -30\), and we need to find \(c\) such that \(a + b + c = 0\).

2. **Set up the equation**: From the identity, we know:
\[ 49 + (-30) + c = 0 \]

3. **Simplify the equation**: Calculate \(49 - 30\):
\[ 19 + c = 0 \]

4. **Solve for \(c\)**: Rearranging gives:
\[ c = -19 \]

5. **Substitute back into the equation**: Now we can substitute \(c\) back into our original equation:
\[ 49^3 - 30^3 + (-19)^3 = 3 \times 49 \times (-30) \times (-19) \]

6. **Calculate the right-hand side**: We can compute:
\[ 3 \times 49 \times 30 \times 19 \]

7. **Final equation**: Thus, we conclude that:
\[ 49^3 - 30^3 + (-19)^3 = 3 \times 49 \times 30 \times 19 \]

### Conclusion:
The blank in the equation can be filled with \(-19\).