If `a+b+c=7, and a^(2)+b^(2)+c^(2)=33`, then what is the value of `a^(3)+b^(3)+c^(3)-3abc`?

Options

A. 257
B. 287
C. 343
D. 175

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 \( a^2 + b^2 + c^2 = 33 \).

### Step-by-step Solution:

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

2. **Substitute known values**:
We have \( a + b + c = 7 \) and \( a^2 + b^2 + c^2 = 33 \). We need to find \( ab + ac + bc \).

3. **Find \( ab + ac + bc \)**:
We can use the square of the sum of \( a, b, c \):
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting the known values:
\[ 7^2 = 33 + 2(ab + ac + bc) \]
This simplifies to:
\[ 49 = 33 + 2(ab + ac + bc) \]
\[ 49 - 33 = 2(ab + ac + bc) \]
\[ 16 = 2(ab + ac + bc) \]
\[ ab + ac + bc = \frac{16}{2} = 8 \]

4. **Substitute back into the identity**:
Now we can substitute \( ab + ac + bc \) back into the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
We already know \( a + b + c = 7 \) and \( a^2 + b^2 + c^2 = 33 \), so we calculate:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 33 - 8 = 25 \]
Thus:
\[ a^3 + b^3 + c^3 - 3abc = 7 \times 25 = 175 \]

5. **Final answer**:
Therefore, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( \boxed{175} \).