Two real numbers $x$ and $y$ satisfy $x-y=4$ and $x^3-y^3=28$.  Compute $xy$.
Solution 1. The first equation gives $x = y+4$. Substituting into the second equation, we get \[(y+4)^3 - y^3 = 28 \implies 12y^2 + 48y + 36 = 0.\]Thus, $y^2 + 4y + 3 = 0$, so $(y+1)(y+3) = 0$. Therefore, either $y=-1$ and $x=y+4=3$, or $y=-3$ and $x=y+4=1$. Either way, $xy = \boxed{-3}$.

Solution 2. The second equation factors via difference of cubes, as \[(x-y)(x^2+xy+y^2) = 28.\]Since $x-y=4$, we have $x^2+xy+y^2=\frac{28}{4} =7$. Now, squaring the first equation, we get $x^2-2xy+y^2=16$. Thus, \[3xy = (x^2+xy+y^2) - (x^2-2xy+y^2) = 7-16=-9,\]so $xy = \frac{-9}{3} = \boxed{-3}$.
-3
Level 1
Intermediate Algebra
Problem: Two real numbers $x$ and $y$ satisfy $x-y=4$ and $x^3-y^3=28$.  Compute $xy$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( xy \), we start with the equations \( x - y = 4 \) and \( x^3 - y^3 = 28 \). 

First, we can use the identity for the difference of cubes: 
\[
x^3 - y^3 = (x - y)(x^2 + xy + y^2).
\]
Substituting \( x - y = 4 \) into this identity gives:
\[
x^3 - y^3 = 4(x^2 + xy + y^2).
\]
Setting this equal to 28, we have:
\[
4(x^2 + xy + y^2) = 28.
\]
Dividing both sides by 4 results in:
\[
x^2 + xy + y^2 = 7.
\]

Next, we can express \( x \) in terms of \( y \) using the first equation:
\[
x = y + 4.
\]
Substituting this expression for \( x \) into the equation \( x^2 + xy + y^2 = 7 \) gives:
\[
(y + 4)^2 + (y + 4)y + y^2 = 7.
\]
Expanding \( (y + 4)^2 \) results in:
\[
y^2 + 8y + 16.
\]
The term \( (y + 4)y \) expands to:
\[
y^2 + 4y.
\]
Combining these, we have:
\[
y^2 + 8y + 16 + y^2 + 4y + y^2 = 7.
\]
This simplifies to:
\[
3y^2 + 12y + 16 = 7.
\]
Subtracting 7 from both sides yields:
\[
3y^2 + 12y + 9 = 0.
\]
Dividing the entire equation by 3 simplifies it to:
\[
y^2 + 4y + 3 = 0.
\]
Factoring this quadratic equation gives:
\[
(y + 1)(y + 3) = 0.
\]
Thus, the solutions for \( y \) are:
\[
y = -1 \quad \text{or} \quad y = -3.
\]

Now, we find the corresponding values of \( x \) for each \( y \):
1. If \( y = -1 \):
   \[
   x = -1 + 4 = 3.
   \]
   Therefore, \( xy = 3 \cdot (-1) = -3 \).

2. If \( y = -3 \):
   \[
   x = -3 + 4 = 1.
   \]
   Thus, \( xy = 1 \cdot (-3) = -3 \).

In both cases, we find that \( xy = -3 \). Therefore, the final answer is \( -3 \).<|END_OF_LATENT|> The final answer is $-3$.