Đáp án:

\(\left( {1; - 2} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}y = m{x^3} - 2\left( {{m^2} + 1} \right){x^2} + 2{m^2} - m\,\,\forall m\\ \Leftrightarrow m{x^3} - 2{m^2}{x^2} - 2{x^2} + 2{m^2} - m - y = 0\,\,\forall m\\ \Leftrightarrow 2{m^2}\left( {1 - {x^2}} \right) + m\left( {{x^3} - 1} \right) - 2{x^2} - y = 0\,\,\forall m\\ \Rightarrow \left\{ \begin{array}{l}1 - {x^2} = 0\\{x^3} - 1 = 0\\ - 2{x^2} - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x =  - 1\end{array} \right.\\x = 1\\ - 2{x^2} - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\ - 2 - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\y =  - 2\end{array} \right.\end{array}\)

Vậy đồ thị hàm số luôn đi qua điểm \(\left( {1; - 2} \right)\) với mọi \(m\).

$\begin{array}{l}y = m{x^3} - 2\left( {{m^2} + 1} \right){x^2} + 2{m^2} - m\,\,\forall m\\ \Leftrightarrow m{x^3} - 2{m^2}{x^2} - 2{x^2} + 2{m^2} - m - y = 0\,\,\forall m\\ \Leftrightarrow 2{m^2}\left( {1 - {x^2}} \right) + m\left( {{x^3} - 1} \right) - 2{x^2} - y = 0\,\,\forall m\\ \Rightarrow \left\{ \begin{array}{l}1 - {x^2} = 0\\{x^3} - 1 = 0\\ - 2{x^2} - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x = - 1\end{array} \right.\\x = 1\\ - 2{x^2} - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\ - 2 - y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\\y = - 2\end{array} \right.\end{array}$