jmc

From $ab+ac+bc=0,$ we get $ab=-ac-bc,$ $ac=-ab-bc,$ and $bc=-ab-ac.$ Then $$\begin{array}{rl}(ab-c)(ac-b)(bc-a)& =(-ac-bc-c)(-ab-bc-b)(-ab-ac-a)\\ & =-abc(a+b+1)(a+c+1)(b+c+1).\end{array}$$Let $s=a+b+c.$ Then $$-abc(a+b+1)(a+c+1)(b+c+1)=-abc(s+1-c)(s+1-b)(s+1-a).$$We know that $a,$ $b,$ and $c$ are the roots of the polynomial $$p(x)=(x-a)(x-b)(x-c).$$Expanding, we get $$p(x)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc.$$We know that $ab+ac+bc=0.$ Also, $abc=(a+b+c+1{)}^2=(s+1{)}^2,$ so $$p(x)=x^3-s{x}^2-(s+1{)}^2.$$Setting $x=s+1,$ we get $$p(s+1)=(s+1{)}^3-s(s+1{)}^2-(s+1{)}^2=0.$$But $$p(s+1)=(s+1-a)(s+1-b)(s+1-c).$$Therefore, $$-abc(s+1-c)(s+1-b)(s+1-a)=0.$$The only possible value of the given expression is $\boxed{0}.$ The triple $(a,b,c)=(1,-2,-2)$ shows that the value of 0 is achievable.