If ` x + y + z = 1, x^(2) + y^(2) + z^(2) = 2 and x^(3) + y^(3) + z^(3) = 3 ` then what is the value of xyz ?

Options

A. `(1)/(3)`
B. `(1)/(6)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To find the value of \( xyz \) given the equations:

1. \( x + y + z = 1 \)
2. \( x^2 + y^2 + z^2 = 2 \)
3. \( x^3 + y^3 + z^3 = 3 \)

we can use the relationships between the sums of powers of the variables and the elementary symmetric sums.

### Step 1: Use the identity for the sum of squares
We know the identity:
\[ x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) \]
Substituting the known values:
\[ 2 = (1)^2 - 2(xy + yz + zx) \]
This simplifies to:
\[ 2 = 1 - 2(xy + yz + zx) \]
Rearranging gives:
\[ 2(xy + yz + zx) = 1 - 2 \]
\[ 2(xy + yz + zx) = -1 \]
Thus:
\[ xy + yz + zx = -\frac{1}{2} \]

### Step 2: Use the identity for the sum of cubes
Next, we use the identity:
\[ x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz \]
Substituting the known values:
\[ 3 = 1 \left(2 - \left(-\frac{1}{2}\right)\right) + 3xyz \]
This simplifies to:
\[ 3 = 1 \left(2 + \frac{1}{2}\right) + 3xyz \]
\[ 3 = 1 \left(\frac{4}{2} + \frac{1}{2}\right) + 3xyz \]
\[ 3 = \frac{5}{2} + 3xyz \]

### Step 3: Solve for \( xyz \)
Rearranging gives:
\[ 3 - \frac{5}{2} = 3xyz \]
Converting \( 3 \) to a fraction:
\[ \frac{6}{2} - \frac{5}{2} = 3xyz \]
\[ \frac{1}{2} = 3xyz \]
Dividing both sides by 3:
\[ xyz = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]

### Final Answer
Thus, the value of \( xyz \) is:
\[ \boxed{\frac{1}{6}} \]