jmc

If $(x-1{)}^2$ is a factor of $x^3+bx+c,$ then the other factor must be $x+c,$ to make the leading and constant coefficients match. Thus, $$(x-1{)}^2(x+c)=x^3+bx+c.$$Expanding, we get $$x^3+(c-2)x^2+(1-2c)x+c=x^3+bx+c.$$Matching coefficients, we get $$\begin{array}{rl}c-2& =0,\\ 1-2c& =b.\end{array}$$Then $c=2,$ so $b=1-2c=-3.$ Thus, $(b,c)=\boxed{(-3,2)}.$