If ` x^(3) - y^(3) = 112 and x - y = 4 ` then what is the value of ` x^(2) + y^(2)` ?

Options

A. 16
B. 20
C. 24
D. 28

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 \) given the equations \( x^3 - y^3 = 112 \) and \( x - y = 4 \).

### Step-by-Step Solution:

1. **Use the identity for the difference of cubes:**
The difference of cubes can be factored as:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
We know \( x - y = 4 \). Therefore, we can substitute this into the equation:
\[ x^3 - y^3 = 4(x^2 + xy + y^2) \]

2. **Substitute the known value into the equation:**
We are given that \( x^3 - y^3 = 112 \). So we can set up the equation:
\[ 112 = 4(x^2 + xy + y^2) \]

3. **Solve for \( x^2 + xy + y^2 \):**
Divide both sides by 4:
\[ x^2 + xy + y^2 = \frac{112}{4} = 28 \]

4. **Use the identity for \( x^2 + y^2 \):**
We know that:
\[ x^2 + y^2 = (x - y)^2 + 2xy \]
We already have \( x - y = 4 \), so:
\[ (x - y)^2 = 4^2 = 16 \]
Thus, we can rewrite the equation as:
\[ x^2 + y^2 = 16 + 2xy \]

5. **Express \( xy \) in terms of known quantities:**
From the equation \( x^2 + xy + y^2 = 28 \), we can substitute \( x^2 + y^2 \):
\[ 16 + 2xy + xy = 28 \]
This simplifies to:
\[ 16 + 3xy = 28 \]
Subtract 16 from both sides:
\[ 3xy = 12 \]
Divide by 3:
\[ xy = 4 \]

6. **Substitute \( xy \) back into the equation for \( x^2 + y^2 \):**
Now substitute \( xy = 4 \) into \( x^2 + y^2 = 16 + 2xy \):
\[ x^2 + y^2 = 16 + 2(4) = 16 + 8 = 24 \]

### Final Answer:
Thus, the value of \( x^2 + y^2 \) is:
\[ \boxed{24} \]