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

1. \( x + y + z = 6 \)
2. \( x^2 + y^2 + z^2 = 20 \)

We can use the identity for the sum of cubes:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

### Step 1: Calculate \( xy + yz + zx \)

From the equation \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \), we can substitute the known values:

\[ (x + y + z)^2 = 6^2 = 36 \]
\[ x^2 + y^2 + z^2 = 20 \]

Now, substituting these into the equation:

\[ 36 = 20 + 2(xy + yz + zx) \]

### Step 2: Solve for \( xy + yz + zx \)

Rearranging the equation gives:

\[ 36 - 20 = 2(xy + yz + zx) \]
\[ 16 = 2(xy + yz + zx) \]
\[ xy + yz + zx = \frac{16}{2} = 8 \]

### Step 3: Substitute into the identity

Now we can substitute \( x + y + z \) and \( xy + yz + zx \) into the identity:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

Calculating \( x^2 + y^2 + z^2 - xy - yz - zx \):

\[ x^2 + y^2 + z^2 - xy - yz - zx = 20 - 8 = 12 \]

### Step 4: Final calculation

Now substituting back into the identity:

\[ x^3 + y^3 + z^3 - 3xyz = (6)(12) = 72 \]

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

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