If `x + y + z = 3` and `xy + yz + zx = -18`, then what is the value of `x^(3)+y^(3)+z^(3)-3xyz = ?`

Options

A. 187
B. 217
C. 191
D. 189

Answer

To solve the problem, we will follow these steps:

1. **Given equations**: We have two equations:
\[ x + y + z = 3 \quad (1) \]
\[ xy + yz + zx = -18 \quad (2) \]

2. **Square the first equation**: We will square equation (1):
\[ (x + y + z)^2 = 3^2 \]
Expanding the left side gives:
\[ x^2 + y^2 + z^2 + 2(xy + yz + zx) = 9 \]

3. **Substituting the value from equation (2)**: We know from equation (2) that \(xy + yz + zx = -18\). Substituting this into the squared equation gives:
\[ x^2 + y^2 + z^2 + 2(-18) = 9 \]
Simplifying this:
\[ x^2 + y^2 + z^2 - 36 = 9 \]
\[ x^2 + y^2 + z^2 = 9 + 36 = 45 \quad (3) \]

4. **Using the identity for cubes**: We will use the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
We already have \(x + y + z = 3\) and \(x^2 + y^2 + z^2 = 45\). We also need to find \(xy + yz + zx\) which we already know is \(-18\).

5. **Calculate \(x^2 + y^2 + z^2 - xy - yz - zx\)**:
\[ x^2 + y^2 + z^2 - (xy + yz + zx) = 45 - (-18) \]
\[ = 45 + 18 = 63 \]

6. **Substituting back into the identity**:
Now substituting everything back into the identity:
\[ x^3 + y^3 + z^3 - 3xyz = 3 \cdot 63 = 189 \]

Thus, the value of \(x^3 + y^3 + z^3 - 3xyz\) is:
\[ \boxed{189} \]