### Instruction: Two real numbers $x$ and $y$ satisfy $x-y=4$ and $x^3-y^3=28$.  Compute $xy$.### Response: I notice that the second equation is a difference of cubes, which can be factored as $(x-y)(x^2+xy+y^2)$.
I also notice that I already know the value of $x-y$, which is 4.
So I can substitute that into the second equation and get $4(x^2+xy+y^2)=28$.
Dividing both sides by 4, I get $x^2+xy+y^2=7$.
Now I want to find a way to use the first equation, $x-y=4$, to simplify this.
One idea is to square both sides of the first equation and see what happens.
If I do that, I get $x^2-2xy+y^2=16$.
I notice that this has some terms in common with the other equation, so I can subtract them and eliminate the squared terms.
If I subtract $x^2-2xy+y^2=16$ from $x^2+xy+y^2=7$, I get $3xy=-9$.
Solving for $xy$, I get $xy=-3$.
So that's my answer.