Đáp án:
$\begin{array}{l}
+ )\left( {9{x^2} - 4} \right)\left( {x + 1} \right) = \left( {3x + 2} \right)\left( {{x^2} - 1} \right)\\
\Rightarrow \left( {9{x^2} - 4} \right)\left( {x + 1} \right) = \left( {3x + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
\Rightarrow \left( {x + 1} \right)\left( {9{x^2} - 4 - \left( {3x + 2} \right)\left( {x - 1} \right)} \right) = 0\\
\Rightarrow \left( {x + 1} \right)\left( {9{x^2} - 4 - 3{x^2} + x + 2} \right) = 0\\
\Rightarrow \left( {x + 1} \right)\left( {6{x^2} + x - 2} \right) = 0\\
\Rightarrow \left( {x + 1} \right)\left( {2x - 1} \right)\left( {3x + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\\
x = \frac{1}{2}\\
x = - \frac{2}{3}
\end{array} \right.\\
+ ){x^3} - 7x + 6 = 0\\
\Rightarrow {x^3} - {x^2} + {x^2} - x - 6x + 6 = 0\\
\Rightarrow \left( {x - 1} \right)\left( {{x^2} + x - 6} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = - 3
\end{array} \right.
\end{array}$
