If `a+b+c = 4` and `ab + bc + ca = 1`, then the value of `a^(3) + b^(3) + c^(3) - 3abc` is :

Options

A. 50
B. 60
C. 52
D. 47

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given that \( a + b + c = 4 \) and \( ab + bc + ca = 1 \).

### Step-by-Step Solution:

1. **Use the identity for \( a^3 + b^3 + c^3 - 3abc \)**:
The formula states:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

2. **Substitute \( a + b + c \)**:
Given \( a + b + c = 4 \), we can substitute this into the formula:
\[ a^3 + b^3 + c^3 - 3abc = 4(a^2 + b^2 + c^2 - ab - ac - bc) \]

3. **Find \( a^2 + b^2 + c^2 \)**:
We know that:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a^2 + b^2 + c^2 = 4^2 - 2 \cdot 1 = 16 - 2 = 14 \]

4. **Substitute \( a^2 + b^2 + c^2 \) into the equation**:
Now we have:
\[ a^3 + b^3 + c^3 - 3abc = 4(14 - 1) \]
Simplifying this gives:
\[ a^3 + b^3 + c^3 - 3abc = 4 \cdot 13 = 52 \]

5. **Final Result**:
Therefore, the value of \( a^3 + b^3 + c^3 - 3abc \) is:
\[ \boxed{52} \]