jmc

Notice that $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=x^3-6x^2+5x+12$. Thus by finding the roots we will determine the set $\{a,b,c\}$. But the roots are $x=-1,3,4$, so we see that $a^3+b^3+c^3=-1+27+64=\boxed{90}$.