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, we have the equations:

1. \( (x + y)^2 = xy + 1 \)
2. \( x^3 - y^3 = 1 \)

We need to find the value of \( x - y \).

### Step 1: Expand the first equation

Starting with the first equation:

\[ (x + y)^2 = xy + 1 \]

Expanding the left side:

\[ x^2 + 2xy + y^2 = xy + 1 \]

### Step 2: Rearrange the equation

Now, rearranging the equation gives:

\[ x^2 + 2xy + y^2 - xy - 1 = 0 \]

This simplifies to:

\[ x^2 + xy + y^2 - 1 = 0 \]

### Step 3: Use the second equation

Now, let's use the second equation:

\[ x^3 - y^3 = 1 \]

Using the identity for the difference of cubes, we can rewrite this as:

\[ (x - y)(x^2 + xy + y^2) = 1 \]

### Step 4: Substitute \( x^2 + xy + y^2 \)

From our rearranged first equation, we know that \( x^2 + xy + y^2 = 1 \). Substituting this into the difference of cubes equation gives:

\[ (x - y)(1) = 1 \]

### Step 5: Solve for \( x - y \)

This simplifies to:

\[ x - y = 1 \]

### Conclusion

Thus, the value of \( x - y \) is:

\[ \boxed{1} \]

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