If `a + b + c = 7 and a b + bc + ca = 16` then the value of `a ^(3) + b ^(3) + c ^(3) - 3 abc ` is

Options

A. 7
B. 16
C. `1//2`
D. 23

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given that \( a + b + c = 7 \) and \( ab + bc + ca = 16 \).

### Step-by-Step Solution:

1. **Use the identity for the sum of cubes**:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
We will denote \( S_1 = a + b + c \) and \( S_2 = ab + ac + bc \).

2. **Substitute the known values**:
From the problem, we have:
\[ S_1 = 7 \quad \text{and} \quad S_2 = 16 \]

3. **Find \( a^2 + b^2 + c^2 \)**:
We can use the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a^2 + b^2 + c^2 = S_1^2 - 2S_2 = 7^2 - 2 \times 16 \]
\[ = 49 - 32 = 17 \]

4. **Calculate \( a^2 + b^2 + c^2 - ab - ac - bc \)**:
Now we can find:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 17 - 16 = 1 \]

5. **Substitute back into the identity**:
Now we substitute back into our original identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
\[ = 7 \times 1 = 7 \]

### Final Answer:
Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( \boxed{7} \).