If `x^(3)-y^(3)=81 and x-y=3,` what is the value of `x^(2)+y^(2)`?

Options

A. 18
B. 21
C. 27
D. 36

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 \) given the equations \( x^3 - y^3 = 81 \) and \( x - y = 3 \).

### Step-by-Step Solution:

1. **Use the identity for the difference of cubes**:
The difference of cubes can be expressed as:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Given \( x^3 - y^3 = 81 \) and \( x - y = 3 \), we can substitute these values into the identity:
\[ 81 = 3(x^2 + xy + y^2) \]

2. **Solve for \( x^2 + xy + y^2 \)**:
Dividing both sides of the equation by 3 gives:
\[ x^2 + xy + y^2 = \frac{81}{3} = 27 \]

3. **Use the identity for \( x^2 + y^2 \)**:
We know that:
\[ x^2 + y^2 = (x - y)^2 + 2xy \]
Substituting \( x - y = 3 \):
\[ x^2 + y^2 = 3^2 + 2xy = 9 + 2xy \]

4. **Find \( xy \)**:
We can express \( x^2 + xy + y^2 \) in terms of \( x^2 + y^2 \):
\[ x^2 + xy + y^2 = x^2 + y^2 + xy \]
Substituting \( x^2 + y^2 = 9 + 2xy \) into the equation:
\[ 27 = (9 + 2xy) + xy \]
Simplifying gives:
\[ 27 = 9 + 3xy \]
Subtracting 9 from both sides:
\[ 18 = 3xy \]
Dividing by 3:
\[ xy = 6 \]

5. **Substitute \( xy \) back into the equation for \( x^2 + y^2 \)**:
Now we can substitute \( xy = 6 \) back into the equation for \( x^2 + y^2 \):
\[ x^2 + y^2 = 9 + 2(6) = 9 + 12 = 21 \]

### Final Answer:
Thus, the value of \( x^2 + y^2 \) is \( \boxed{21} \).