Đáp án:

\(m = \pm {1 \over 2}\)

Giải thích các bước giải:

\(\eqalign{ & y = - {x^3} + 3{x^2} + 3\left( {{m^2} - 1} \right)x - 3{m^2} - 1 \cr & y' = - 3{x^2} + 6x + 3\left( {{m^2} - 1} \right) = 0 \cr & \Leftrightarrow {x^2} - 2x - {m^2} + 1 = 0\,\,\left( * \right) \cr & Ham\,\,so\,\,co\,\,cuc\,\,dai,\,\,cuc\,\,tieu \cr & \Leftrightarrow \left( * \right)\,\,co\,\,2\,\,nghiem\,\,pb \cr & \Rightarrow \Delta ' = 1 + {m^2} - 1 = {m^2} > 0 \Leftrightarrow m \ne 0 \cr & \Rightarrow \left[ \matrix{ {x_1} = 1 + m \Rightarrow {y_1} = 2{m^3} - 2 \hfill \cr {x_2} = 1 - m \Rightarrow {y_2} = - 2{m^3} - 2 \hfill \cr} \right. \cr & \Rightarrow A\left( {1 + m;2{m^3} - 2} \right);\,\,B\left( {1 - m; - 2{m^3} - 2} \right) \cr & A,\,B\,\,cach\,\,deu\,\,O \Rightarrow OA = OB \cr & \Rightarrow O{A^2} = O{B^2} \cr & \Rightarrow {\left( {1 + m} \right)^2} + {\left( {2{m^3} - 2} \right)^2} = {\left( {1 - m} \right)^2} + {\left( {2{m^3} + 2} \right)^2} \cr & \Leftrightarrow 1 + 2m + {m^2} + 4{m^6} - 8{m^3} + 4 = 1 - 2m + {m^2} + 4{m^6} + 8{m^3} + 4 \cr & \Leftrightarrow 4m - 16{m^3} = 0 \cr & \Leftrightarrow 4m\left( {1 - 4{m^2}} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ m = 0\,\,\left( {loai} \right) \hfill \cr m = \pm {1 \over 2} \hfill \cr} \right. \cr & Vay\,\,m = \pm {1 \over 2}. \cr} \)