If `x+y+z=0`, then find `(x+y)^(3)+(y+z)^(3)+(z+x)^(3)`.

Answer

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

**Given:**
\( x + y + z = 0 \)

We need to find the value of:
\( (x+y)^3 + (y+z)^3 + (z+x)^3 \)

### Step 1: Express \( (x+y) \), \( (y+z) \), and \( (z+x) \) in terms of \( z \), \( x \), and \( y \) respectively.

From the equation \( x + y + z = 0 \), we can rearrange it to express each pair in terms of the third variable:

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

### Step 2: Substitute these expressions into the original equation.

Now we substitute these values into the expression we need to evaluate:

\[ (x+y)^3 + (y+z)^3 + (z+x)^3 = (-z)^3 + (-x)^3 + (-y)^3 \]

### Step 3: Simplify the expression.

Using the property of cubes, we know that:

\[ (-a)^3 = -a^3 \]

Thus, we can rewrite the expression as:

\[ (-z)^3 + (-x)^3 + (-y)^3 = -z^3 - x^3 - y^3 \]

This simplifies to:

\[ -(x^3 + y^3 + z^3) \]

### Step 4: Use the identity for the sum of cubes.

There is a known identity that states:

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

Since \( x + y + z = 0 \), we can simplify this identity:

\[ x^3 + y^3 + z^3 = 3xyz \]

### Step 5: Substitute back into our expression.

Now we substitute this back into our expression:

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

### Final Answer:

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

\[ \boxed{-3xyz} \]

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