jmc

By Vieta's formulas, $$\begin{array}{rl}a+b+c& =0,\\ ab+ac+bc& =-1,\\ abc& =1.\end{array}$$Then $$\begin{array}{rl}a(b-c{)}^2+b(c-a{)}^2+c(a-b{)}^2& =a(b^2-2bc+c^2)+b(c^2-2ac+a^2)+c(a^2-2ab+b^2)\\ & =(a{b}^2-2abc+a{c}^2)+(b{c}^2-2abc+b{a}^2)+(c{a}^2-2abc+c{b}^2)\\ & =(a{b}^2-2+a{c}^2)+(b{c}^2-2+b{a}^2)+(c{a}^2-2+c{b}^2)\\ & =a{b}^2+a{c}^2+b{c}^2+b{a}^2+c{a}^2+c{b}^2-6\\ & =a^2(b+c)+b^2(a+c)+c^2(a+b)-6.\end{array}$$From $a+b+c=0,$ $b+c=-a.$ Simillarly, $a+c=-b$ and $a+b=-c,$ so $$a^2(b+c)+b^2(a+c)+c^2(a+b)-6=-a^3-b^3-c^3-6.$$Since $a$ is a root of $x^3-x-1=0,$ $a^3-a-1=0,$ so $-a^3=-a-1.$ Similarly, $-b^3=-b-1$ and $-c^3=-c-1,$ so $$\begin{array}{rl}-a^3-b^3-c^3-6& =(-a-1)+(-b-1)+(-c-1)-6\\ & =-(a+b+c)-9\\ & =\boxed{-9}.\end{array}$$