Find the value of `a^(3)+b^(3)+c^(3)-3abc`
if `a+b+c=8` and `ab+bc+ca=19`.

Answer

To find the value of \( a^3 + b^3 + c^3 - 3abc \) given that \( a + b + c = 8 \) and \( ab + bc + ca = 19 \), we can use the algebraic identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step-by-Step Solution:

1. **Identify the given values**:
- \( a + b + c = 8 \)
- \( ab + ac + bc = 19 \)

2. **Find \( a^2 + b^2 + c^2 \)** using the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substitute the known values:
\[ a^2 + b^2 + c^2 = 8^2 - 2 \times 19 \]
\[ = 64 - 38 = 26 \]

3. **Calculate \( a^2 + b^2 + c^2 - ab - ac - bc \)**:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 26 - 19 = 7 \]

4. **Substitute back into the identity**:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
\[ = 8 \times 7 = 56 \]

### Final Answer:
Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( \boxed{56} \).