If `(x + y) ^(2) = xy +1 and x ^(3) -y ^(3) = 1,` what is the value of `(x-y)` ?

Options

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

Answer

To solve the problem step by step, we will use the given equations and some algebraic identities.

### Step 1: Expand the first equation
We start with the equation:
\[ (x + y)^2 = xy + 1 \]
Expanding the left side, we get:
\[ x^2 + 2xy + y^2 = xy + 1 \]
Now, we can rearrange this equation:
\[ x^2 + y^2 + 2xy - xy - 1 = 0 \]
This simplifies to:
\[ x^2 + y^2 + xy - 1 = 0 \quad \text{(Equation 1)} \]

### Step 2: Use the second equation
The second equation given is:
\[ x^3 - y^3 = 1 \]
We can use the identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Setting this equal to 1, we have:
\[ (x - y)(x^2 + xy + y^2) = 1 \quad \text{(Equation 2)} \]

### Step 3: Express \(x^2 + xy + y^2\) in terms of Equation 1
From Equation 1, we know:
\[ x^2 + y^2 = 1 - xy \]
Now, we can express \(x^2 + xy + y^2\):
\[ x^2 + xy + y^2 = (x^2 + y^2) + xy = (1 - xy) + xy = 1 \]

### Step 4: Substitute back into Equation 2
Now substituting \(x^2 + xy + y^2 = 1\) into Equation 2:
\[ (x - y)(1) = 1 \]
This simplifies to:
\[ x - y = 1 \]

### Final Answer
Thus, the value of \(x - y\) is:
\[ \boxed{1} \]