If `a+b+c=8 and ab +bc +ca =12`, then `a^(3) +b^(3) +c^(3) -3abc` is equal to :

Options

A. a) 144
B. b) 400
C. c) 224
D. d) 192

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given the equations \( a + b + c = 8 \) and \( ab + bc + ca = 12 \).

### Step-by-Step Solution:

1. **Use the identity for cubes**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
This means we need to calculate \( a^2 + b^2 + c^2 \) first.

2. **Calculate \( a^2 + b^2 + c^2 \)**:
We know from the first equation that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting the known values:
\[ 8^2 = a^2 + b^2 + c^2 + 2(12) \]
This simplifies to:
\[ 64 = a^2 + b^2 + c^2 + 24 \]
Rearranging gives:
\[ a^2 + b^2 + c^2 = 64 - 24 = 40 \]

3. **Substitute values into the identity**:
Now we substitute \( a + b + c = 8 \) and \( a^2 + b^2 + c^2 = 40 \) into the identity:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 40 - 12 = 28 \]

4. **Calculate \( a^3 + b^3 + c^3 - 3abc \)**:
Now we can substitute back into our identity:
\[ a^3 + b^3 + c^3 - 3abc = (8)(28) = 224 \]

Thus, the final answer is:
\[ \boxed{224} \]