What is the expression `(x+y)^(-1) (x^(-1) +y^(-1)) (xy^(-1) +x^(-1)y)^(-1)` equal to:

Options

A. `x + y`
B. `(x^(2) +y^(2))^(-1)`
C. xy
D. `x^(2) + y^(2)`

Answer

To solve the expression \((x+y)^{-1} (x^{-1} + y^{-1}) (xy^{-1} + x^{-1}y)^{-1}\), we will simplify it step by step.

### Step 1: Rewrite the expression using positive exponents
We start with the expression:
\[ (x+y)^{-1} (x^{-1} + y^{-1}) (xy^{-1} + x^{-1}y)^{-1} \]
Using the property that \(a^{-1} = \frac{1}{a}\), we can rewrite the expression as:
\[ \frac{1}{x+y} \left( \frac{1}{x} + \frac{1}{y} \right) \left( \frac{1}{xy} + \frac{1}{xy} \right)^{-1} \]

### Step 2: Simplify \(x^{-1} + y^{-1}\)
The term \(x^{-1} + y^{-1}\) can be rewritten as:
\[ \frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy} = \frac{x+y}{xy} \]

### Step 3: Simplify \(xy^{-1} + x^{-1}y\)
Now, we simplify \(xy^{-1} + x^{-1}y\):
\[ xy^{-1} + x^{-1}y = \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \]
Thus, the inverse is:
\[ (xy^{-1} + x^{-1}y)^{-1} = \frac{xy}{x^2 + y^2} \]

### Step 4: Substitute back into the expression
Now substituting back into our expression, we have:
\[ \frac{1}{x+y} \cdot \frac{x+y}{xy} \cdot \frac{xy}{x^2 + y^2} \]

### Step 5: Cancel out common terms
Notice that \(x+y\) in the numerator and denominator cancels out:
\[ = \frac{1}{1} \cdot \frac{1}{x^2 + y^2} = \frac{1}{x^2 + y^2} \]

### Final Result
Thus, the expression simplifies to:
\[ \frac{1}{x^2 + y^2} \]