jmc

We recognize part of the expansion of $(x-1{)}^4$ on the left-hand side. Adding $1$ to both sides, we have $$x^4-4x^3+6x^2-4x+1=2006,$$which means $(x-1{)}^4=2006.$ Therefore, $$x-1=\sqrt[4]{2006},i\sqrt[4]{2006},-\sqrt[4]{2006},-i\sqrt[4]{2006}.$$Since we want the nonreal roots, we only consider the roots $$x=1\pm i\sqrt[4]{2006}.$$The product of these roots is $$P=(1+i\sqrt[4]{2006})(1-i\sqrt[4]{2006})=\boxed{1+\sqrt{2006}}.$$