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 the expression:

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

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

### Step-by-Step Solution:

1. **Use the identity for cubes**:
From the identity \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz)\), and since \(x + y + z = 0\), we have:
\[ x^3 + y^3 + z^3 = 3xyz \]
This condition is satisfied.

2. **Substituting \(z\)**:
Since \(x + y + z = 0\), we can express \(z\) as:
\[ z = - (x + y) \]

3. **Substituting in the expression**:
Now, we substitute \(z\) in the expression we need to evaluate:
\[ \frac{(x+y)^2}{xy} = \frac{(y - (x+y))^2}{y(- (x+y))} + \frac{(- (x+y) + x)^2}{(- (x+y))x} \]

4. **Simplifying each term**:
- The left-hand side becomes:
\[ \frac{(x+y)^2}{xy} \]
- The first term on the right-hand side becomes:
\[ \frac{(-x)^2}{-xy} = \frac{x^2}{-xy} = -\frac{x}{y} \]
- The second term on the right-hand side becomes:
\[ \frac{(-y)^2}{(- (x+y))x} = \frac{y^2}{(- (x+y))x} = -\frac{y}{x+y} \]

5. **Combining the right-hand side**:
Now, we combine the right-hand side:
\[ -\frac{x}{y} - \frac{y}{x+y} \]

6. **Finding a common denominator**:
The common denominator for the right-hand side is \(xy\):
\[ -\frac{x^2 + y^2}{xy} \]

7. **Setting the two sides equal**:
Now we set the left-hand side equal to the right-hand side:
\[ \frac{(x+y)^2}{xy} = -\frac{x^2 + y^2}{xy} \]

8. **Cross-multiplying**:
Cross-multiplying gives:
\[ (x+y)^2 = - (x^2 + y^2) \]
This implies:
\[ x^2 + 2xy + y^2 = -x^2 - y^2 \]
Rearranging gives:
\[ 2xy + 2x^2 + 2y^2 = 0 \]
Thus:
\[ xy + x^2 + y^2 = 0 \]

9. **Conclusion**:
Since \(x + y + z = 0\) implies \(x + y = -z\), we can conclude that the expression holds true and simplifies to \(3\).

### Final Answer:
The value of the expression is \(3\).