If `x^(3) + y^(3) + z^(3) = 3xyz and x + y + z = 0`, find the value of :
`((x+y)^(2))/(xy) + ((y+z)^(2))/(yz) + ((z+x)^(2))/(zx)`

Answer

To solve the problem, we need to find the value of:

\[ \frac{(x+y)^2}{xy} + \frac{(y+z)^2}{yz} + \frac{(z+x)^2}{zx} \]

given that \(x^3 + y^3 + z^3 = 3xyz\) and \(x + y + z = 0\).

### Step 1: Use the identity for \(x^3 + y^3 + z^3\)

From the identity, we know that:

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

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

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

This confirms the given condition.

### Step 2: Express \(x+y\), \(y+z\), and \(z+x\)

From the equation \(x + y + z = 0\), we can express:

- \(x + y = -z\)
- \(y + z = -x\)
- \(z + x = -y\)

### Step 3: Substitute these expressions into the original equation

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

\[ \frac{(x+y)^2}{xy} + \frac{(y+z)^2}{yz} + \frac{(z+x)^2}{zx} \]

This becomes:

\[ \frac{(-z)^2}{xy} + \frac{(-x)^2}{yz} + \frac{(-y)^2}{zx} \]

Which simplifies to:

\[ \frac{z^2}{xy} + \frac{x^2}{yz} + \frac{y^2}{zx} \]

### Step 4: Find a common denominator

The common denominator for the three fractions is \(xyz\). Thus, we can rewrite the expression as:

\[ \frac{z^2 \cdot z}{xyz} + \frac{x^2 \cdot x}{xyz} + \frac{y^2 \cdot y}{xyz} \]

This simplifies to:

\[ \frac{z^3 + x^3 + y^3}{xyz} \]

### Step 5: Use the identity again

From the condition \(x^3 + y^3 + z^3 = 3xyz\), we substitute this into our expression:

\[ \frac{3xyz}{xyz} = 3 \]

### Conclusion

Thus, the value of the expression is:

\[ \boxed{3} \]