If `x+y+z=9` and `xy+yz+zx=23`, the value of `(x^3+y^3+z^3-3xyz)=? `

Options

A. 108
B. 207
C. 669
D. 729

Explanation

To solve the problem, we need to find the value of \( x^3 + y^3 + z^3 - 3xyz \) given the equations \( x + y + z = 9 \) and \( xy + yz + zx = 23 \).

### Step-by-Step Solution:

1. **Use the identity for \( x^3 + y^3 + z^3 - 3xyz \)**:
The formula we will use is:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z) \left( (x + y + z)^2 - 3(xy + yz + zx) \right) \]

2. **Substitute the known values**:
We know:
\[ x + y + z = 9 \quad \text{and} \quad xy + yz + zx = 23 \]
Substitute these values into the formula:
\[ x^3 + y^3 + z^3 - 3xyz = 9 \left( 9^2 - 3 \times 23 \right) \]

3. **Calculate \( 9^2 \)**:
\[ 9^2 = 81 \]

4. **Calculate \( 3 \times 23 \)**:
\[ 3 \times 23 = 69 \]

5. **Substitute these results back into the equation**:
\[ x^3 + y^3 + z^3 - 3xyz = 9 \left( 81 - 69 \right) \]

6. **Calculate \( 81 - 69 \)**:
\[ 81 - 69 = 12 \]

7. **Multiply by 9**:
\[ x^3 + y^3 + z^3 - 3xyz = 9 \times 12 = 108 \]

### Final Answer:
Thus, the value of \( x^3 + y^3 + z^3 - 3xyz \) is \( \boxed{108} \).