jmc

If $r$ is a root of $f(x)=0$, then $r^3+r^2+2r+3=0$. Rearranging, we have $$r^3+2r=-r^2-3,$$and squaring this equation gives $$r^6+4r^4+4r^2=r^4+6r^2+9,$$or $$r^6+3r^4-2r^2-9=0.$$Rewriting this equation in the form $(r^2{)}^3+3(r^2{)}^2-2r^2-9=0$, we see that the polynomial $x^3+3x^2-2x-9$ has $r^2$ as a root, so three of its roots are the squares of the roots of $f(x)$. But this polynomial is cubic, so these are its only roots. Thus, $g(x)=x^3+3x^2-2x-9$, and so $(b,c,d)=\boxed{(3,-2,-9)}$.