Question:
If $x-y=6$ and $x^2+y^2=24$, find $x^3-y^3$.

Answer:
First, we note  \[x^3-y^3 = (x-y)(x^2 +xy +y^2) = 6(24+xy),\] so we just need to find $xy$ now.  Squaring both sides of $x-y=6$ gives  $$x^2 - 2xy + y^2 = 36.$$ Since $x^2 + y^2 = 24$, we have $24-2xy = 36$, so $xy = -6$, from which we have  \[x^3-y^3 = 6(24 +xy) = 6(24 - 6) = 6(18) = \boxed{108}.\]