If `xy + yz + zx = 1` then `(sum (x + y)/(1 - xy)=`

Options

A. `(4)/(xyz)`
B. `(1)/(xyz)`
C. xyz
D. `x+ y+ z`

Answer

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

**Given:**
\[ xy + yz + zx = 1 \]

We need to evaluate the expression:
\[ \sum \frac{x+y}{1-xy} \]

### Step 1: Rewrite the expression
We can rewrite the expression using the given condition. The expression can be broken down as follows:

\[ \frac{x+y}{1-xy} = \frac{x+y}{(1 - (xy + yz + zx)) + (yz + zx)} = \frac{x+y}{1 - xy - (yz + zx)} \]

### Step 2: Substitute the condition
Since we know that \( xy + yz + zx = 1 \), we can substitute \( 1 \) for \( xy + yz + zx \):

\[ 1 - xy = 1 - xy \]

Thus, we can rewrite the expression as:

\[ \frac{x+y}{1-xy} \]

### Step 3: Simplify the expression
Now, we can simplify the expression further. We can express \( \frac{x+y}{1-xy} \) in terms of \( z \):

\[ \frac{x+y}{1-xy} = \frac{x+y}{z} \]

### Step 4: Sum the expressions
Now we need to sum this expression for all cyclic permutations of \( x, y, z \):

\[ \sum \frac{x+y}{1-xy} = \frac{x+y}{z} + \frac{y+z}{x} + \frac{z+x}{y} \]

### Step 5: Combine the fractions
We can combine these fractions:

\[ = \frac{(x+y)z + (y+z)x + (z+x)y}{xyz} \]

### Step 6: Expand the numerator
Expanding the numerator gives us:

\[ = \frac{(xz + yz + yx + zx + zy + xy)}{xyz} \]

### Step 7: Use the original condition
Using the original condition \( xy + yz + zx = 1 \):

\[ = \frac{1 + xy + yz + zx}{xyz} = \frac{1 + 1}{xyz} = \frac{2}{xyz} \]

### Final Result
Thus, the final result of the summation is:

\[ \sum \frac{x+y}{1-xy} = \frac{2}{xyz} \]

### Conclusion
The answer to the expression is \( \frac{2}{xyz} \).

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