Đáp án:
$\begin{array}{l}
{x^3} - \left( {m + 2} \right)\left( {x + 2} \right) + 8 = 0\\
\Rightarrow \left( {{x^3} + 8} \right) - \left( {m + 2} \right)\left( {x + 2} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - \left( {m + 2} \right)\left( {x + 2} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^2} - 2x + 4 - m - 2} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^2} - 2x + 2 - m} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\\
{x^2} - 2x + 2 - m = 0\left( * \right)
\end{array} \right.
\end{array}$
Để pt có 3 nghiệm pb thì pt (*) có 2 nghiệm pb khác -2
$\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
4 - 2.\left( { - 2} \right) + 2 - m \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
10 - m \ne 0\\
1 - \left( {2 - m} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 10\\
1 - 2 + m > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 10\\
m > 1
\end{array} \right.
\end{array}$
