If ` (1)/( x + y) = (1)/(x) + (1)/(y) ( x ne 0, y ne 0, x ne y)` then the value of ` x^(3) - y^(3)` is

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the equation \( \frac{1}{x+y} = \frac{1}{x} + \frac{1}{y} \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \frac{1}{x+y} = \frac{1}{x} + \frac{1}{y} \]

### Step 2: Find a common denominator on the right side
The right side can be combined using the least common multiple (LCM) of \( x \) and \( y \):
\[ \frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy} = \frac{x+y}{xy} \]

### Step 3: Set the fractions equal
Now we can equate the two sides:
\[ \frac{1}{x+y} = \frac{x+y}{xy} \]

### Step 4: Cross-multiply
Cross-multiply to eliminate the fractions:
\[ 1 \cdot xy = (x+y)(x+y) \]
This simplifies to:
\[ xy = (x+y)^2 \]

### Step 5: Expand the right side
Expanding the right side gives:
\[ xy = x^2 + 2xy + y^2 \]

### Step 6: Rearrange the equation
Rearranging the equation results in:
\[ 0 = x^2 + y^2 + xy - xy \]
This simplifies to:
\[ x^2 + y^2 - xy = 0 \]

### Step 7: Recognize the identity
The equation \( x^2 + y^2 - xy = 0 \) can be rewritten using the identity for the difference of cubes:
\[ x^3 - y^3 = (x-y)(x^2 + xy + y^2) \]

### Step 8: Substitute the known values
From our previous step, we know:
\[ x^2 + y^2 = xy \]
Thus:
\[ x^2 + xy + y^2 = xy + xy = 2xy \]

### Step 9: Substitute back into the identity
Now substituting back into the identity gives:
\[ x^3 - y^3 = (x-y)(2xy) \]

### Step 10: Evaluate the expression
Since \( x \neq y \), we can conclude that \( x - y \) is not zero. However, from our earlier equation \( x^2 + y^2 - xy = 0 \), we can also conclude that:
\[ x^2 + y^2 = xy \implies 0 = 0 \]
Thus, \( x^3 - y^3 = 0 \).

### Final Answer
Therefore, the value of \( x^3 - y^3 \) is:
\[ \boxed{0} \]