If `a + b + c = 11` and `ab + bc + ca = 38`, then `a^(3) + b^(3) + c^(3) - 3abc` is equal to :

Options

A. 44
B. 77
C. 55
D. 66

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given the equations \( a + b + c = 11 \) and \( ab + bc + ca = 38 \).

### Step-by-Step Solution:

1. **Use the identity for cubes**:
We know the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
We need to find \( a^2 + b^2 + c^2 \) first.

2. **Find \( a^2 + b^2 + c^2 \)**:
We can use the square of the sum of \( a, b, c \):
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]
Substituting the known values:
\[ 11^2 = a^2 + b^2 + c^2 + 2 \cdot 38 \]
This simplifies to:
\[ 121 = a^2 + b^2 + c^2 + 76 \]
Rearranging gives:
\[ a^2 + b^2 + c^2 = 121 - 76 = 45 \]

3. **Substitute into the identity**:
Now we have \( a + b + c = 11 \) and \( a^2 + b^2 + c^2 = 45 \). We can substitute these into the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]
We need to find \( a^2 + b^2 + c^2 - ab - bc - ca \):
\[ a^2 + b^2 + c^2 - (ab + bc + ca) = 45 - 38 = 7 \]

4. **Calculate \( a^3 + b^3 + c^3 - 3abc \)**:
Now substituting back into the identity:
\[ a^3 + b^3 + c^3 - 3abc = 11 \cdot 7 = 77 \]

Thus, the final answer is:
\[ \boxed{77} \]