If `x + y + z = 6 and x^2 + y^2 + z^2= 20` then the value of `x^3 + y^3 + z^3 - 3xyz` is

Options

A. 64
B. 70
C. 72
D. 76

Answer

To solve the problem, we need to find the value of \( x^3 + y^3 + z^3 - 3xyz \) given the equations \( x + y + z = 6 \) and \( x^2 + y^2 + z^2 = 20 \).

### Step-by-step Solution:

1. **Use the identity for the sum of cubes**:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

2. **Substitute the known values**:
We know that \( x + y + z = 6 \) and \( x^2 + y^2 + z^2 = 20 \). We need to find \( xy + yz + zx \).

3. **Use the identity relating sums and squares**:
The identity states:
\[ x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) \]
Substituting the known values:
\[ 20 = 6^2 - 2(xy + yz + zx) \]
This simplifies to:
\[ 20 = 36 - 2(xy + yz + zx) \]

4. **Rearranging the equation**:
\[ 2(xy + yz + zx) = 36 - 20 \]
\[ 2(xy + yz + zx) = 16 \]
\[ xy + yz + zx = \frac{16}{2} = 8 \]

5. **Substituting back into the identity**:
Now we can substitute \( xy + yz + zx \) back into the identity:
\[ x^2 + y^2 + z^2 - xy - yz - zx = 20 - 8 = 12 \]

6. **Final calculation**:
Now substituting everything back into the sum of cubes identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
\[ = 6 \times 12 = 72 \]

Thus, the value of \( x^3 + y^3 + z^3 - 3xyz \) is **72**.