jmc

Notice that $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=x^3-x^2-4x+4$. Thus by finding the roots of this polynomial, we will determine the set $\{a,b,c\}$. But these are seen by factoring to be $x=1,2,-2$, so we see that $a^3+b^3+c^3=1+8-8=\boxed{1}$.