### Instruction: If $x-y=6$ and $x^2+y^2=24$, find $x^3-y^3$.### Response: I notice that the expression $x^3-y^3$ can be factored as $(x-y)(x^2+xy+y^2)$.
This might be helpful, because I already know the value of $x-y$ and I can use the given equations to find $x^2+xy+y^2$.
To do that, I can square the first equation and subtract the second equation from it.
Squaring the first equation gives $x^2-2xy+y^2=36$, and subtracting the second equation gives $-2xy=12$.
Dividing both sides by -2, I get $xy=-6$.
Now I can plug in $x-y=6$ and $xy=-6$ into the expression $x^2+xy+y^2$.
This gives $x^2+xy+y^2=(x-y)^2+3xy=6^2+3(-6)=36-18=18$.
Therefore, $x^3-y^3=(x-y)(x^2+xy+y^2)=6(18)=108$.