If `(x+y+z)=6 and (xy+yz+zx)=11`, then the value of `(x^(3)+y^(3)+z^(3)-3xyz)` is :

Options

A. 81
B. 54
C. 18
D. none of these

Answer

To solve the problem, we need to find the value of \( x^3 + y^3 + z^3 - 3xyz \) given that \( x + y + z = 6 \) and \( xy + yz + zx = 11 \).

### Step-by-Step Solution:

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

2. **Substitute the known values**:
We know \( x + y + z = 6 \) and \( xy + yz + zx = 11 \). We need to find \( x^2 + y^2 + z^2 \).

3. **Find \( x^2 + y^2 + z^2 \)**:
We can find \( x^2 + y^2 + z^2 \) using the square of the sum:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \]
Substituting the known values:
\[ 6^2 = x^2 + y^2 + z^2 + 2 \cdot 11 \]
This simplifies to:
\[ 36 = x^2 + y^2 + z^2 + 22 \]
Therefore:
\[ x^2 + y^2 + z^2 = 36 - 22 = 14 \]

4. **Substitute back into the identity**:
Now we can substitute \( x^2 + y^2 + z^2 \) back into the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z) \left( x^2 + y^2 + z^2 - xy - yz - zx \right) \]
We already have:
- \( x + y + z = 6 \)
- \( x^2 + y^2 + z^2 = 14 \)
- \( xy + yz + zx = 11 \)

Thus:
\[ x^2 + y^2 + z^2 - xy - yz - zx = 14 - 11 = 3 \]

5. **Calculate \( x^3 + y^3 + z^3 - 3xyz \)**:
Now substituting everything back:
\[ x^3 + y^3 + z^3 - 3xyz = 6 \cdot 3 = 18 \]

### Final Answer:
The value of \( x^3 + y^3 + z^3 - 3xyz \) is \( \boxed{18} \).