Đáp án:
$\begin{array}{l}
a){\left( {x - 1} \right)^3} - \left( {x - 3} \right)\left( {x + 3} \right) = {x^2}\left( {x + 3} \right) + 8\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 - {x^2} + 9 = {x^3} + 3{x^2} + 8\\
\Leftrightarrow {x^3} - 4{x^2} + 3x + 8 = {x^3} + 3{x^2} + 8\\
\Leftrightarrow 7{x^2} - 3x = 0\\
\Leftrightarrow x\left( {7x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{3}{7}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{3}{7}\\
b){\left( {2x + 3} \right)^2} = \left( {3x - 4} \right)\left( {3x + 4} \right) + 25\\
\Leftrightarrow 4{x^2} + 12x + 9 = 9{x^2} - 16 + 25\\
\Leftrightarrow 5{x^2} - 12x = 0\\
\Leftrightarrow x\left( {5x - 12} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{12}}{5}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{{12}}{5}\\
c)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - {\left( {x - 2} \right)^3} = 6x\left( {x - 1} \right)\\
\Leftrightarrow {x^3} + 27 - \left( {{x^3} - 6{x^2} + 12x - 8} \right) = 6{x^2} - 6x\\
\Leftrightarrow {x^3} + 27 - {x^3} + 6{x^2} - 12x + 8 = 6{x^2} - 6x\\
\Leftrightarrow 6x = 35\\
\Leftrightarrow x = \dfrac{{35}}{6}\\
Vậy\,x = \dfrac{{35}}{6}
\end{array}$
