If $x-y=6$ and $x^2+y^2=24$, find $x^3-y^3$.
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 $xy=-6$. Therefore, $x^3-y^3=(x-y)(x^2+xy+y^2)=6(18)=108$. 