jmc

If $p$ is a root of $x^3-x^2+x-2=0$, then $p^3-p^2+p-2=0$, or $$p^3=p^2-p+2.$$Similarly, $q^3=q^2-q+2$, and $r^3=r^2-r+2$, so $$p^3+q^3+r^3=(p^2+q^2+r^2)-(p+q+r)+6.$$By Vieta's formulas, $p+q+r=1$, $pq+pr+qr=1$, and $pqr=2$. Squaring the equation $p+q+r=1$, we get $$p^2+q^2+r^2+2pq+2pr+2qr=1.$$Subtracting $2pq+2pr+2qr=2$, we get $$p^2+q^2+r^2=-1.$$Therefore, $p^3+q^3+r^3=(p^2+q^2+r^2)-(p+q+r)+6=(-1)-1+6=\boxed{4}$.