If (x+y+z)=0 then the value of `(x^3+y^3+z^3)` is:

Options

A. `3x^3y^3z^3`
B. 3xyz
C. xyz
D. 3(x+y+z)

Answer

To solve the problem, we start with the given equation:

**Step 1:** Given that \( x + y + z = 0 \).

**Step 2:** We 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 3:** Substitute \( x + y + z = 0 \) into the identity:
\[ x^3 + y^3 + z^3 - 3xyz = 0 \cdot (x^2 + y^2 + z^2 - xy - yz - zx) \]

**Step 4:** Since anything multiplied by zero is zero, we have:
\[ x^3 + y^3 + z^3 - 3xyz = 0 \]

**Step 5:** Rearranging the equation gives us:
\[ x^3 + y^3 + z^3 = 3xyz \]

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

### Final Answer:
\[ x^3 + y^3 + z^3 = 3xyz \]

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