Đáp án:
$\begin{array}{l}
{x^2} - 2mx + {m^2} - 1 = 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = {m^2} - 1
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
x_1^2 - 2m{x_1} + {m^2} - 1 = 0\\
x_2^2 - 2m{x_2} + {m^2} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x_1^2 - 2m{x_1} + {m^2} = 1\\
x_2^2 - 2m{x_2} + {m^2} = 1
\end{array} \right.\\
Goi:{Y^2} - S.Y + P = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{Y_1} = x_1^3 - 2mx_1^2 + {m^2}{x_1} - 2\\
{Y_2} = x_2^3 - 2mx_2^2 + {m^2}{x_2} - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{Y_1} = {x_1}\left( {x_1^2 - 2m{x_1} + {m^2}} \right) - 2\\
{Y_2} = {x_2}\left( {x_2^2 - 2m{x_2} + {m^2}} \right) - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{Y_1} = {x_1}.1 - 2\\
{Y_2} = {x_2}.1 - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{Y_1} = {x_1} - 2\\
{Y_2} = {x_2} - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
S = {Y_1} + {Y_2} = {x_1} + {x_2} - 4\\
P = {Y_1}.{Y_2} = \left( {{x_1} - 2} \right).\left( {{x_2} - 2} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
S = 2m - 4\\
P = {x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
S = 2m - 4\\
P = {m^2} - 1 - 2.2m + 4 = {m^2} - 4m + 3
\end{array} \right.\\
Vậy\,{Y^2} - \left( {2m - 4} \right).Y + {m^2} - 4m + 3 = 0
\end{array}$
