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

Options

A. xyz
B. `3`
C. `3` xyz
D. `1`

Answer

To solve the given problem, we start with the equations provided:

1. \( x^3 + y^3 + z^3 = 3xyz \)
2. \( x + y + z = 0 \)

We need to find the value of the expression:

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

### Step 1: Substitute \( z \) in terms of \( x \) and \( y \)

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

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

### Step 2: Substitute \( z \) into the expression

Now, we substitute \( z \) into the expression we need to evaluate:

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

This simplifies to:

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

### Step 3: Simplify each term

1. The first term remains:
\[ \frac{(x+y)^2}{xy} \]

2. The second term simplifies to:
\[ \frac{x^2}{-y(x+y)} = -\frac{x^2}{y(x+y)} \]

3. The third term simplifies to:
\[ \frac{y^2}{-x(x+y)} = -\frac{y^2}{x(x+y)} \]

### Step 4: Combine the terms

Now, we can combine the terms:

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

To combine these fractions, we find a common denominator, which is \( xy(x+y) \):

\[ \frac{(x+y)^3 - x^2x - y^2y}{xy(x+y)} \]

### Step 5: Simplify the numerator

The numerator simplifies to:

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

Using the identity \( x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz) \), and since \( x+y+z = 0 \):

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

Thus:

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

### Step 6: Substitute back into the expression

Now substituting back, we have:

\[ \frac{-3xyz}{xy(x+y)} \]

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

\[ = \frac{-3xyz}{xy(-z)} = 3 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{3} \]