Đáp án:
$\begin{array}{l}
a)\left( {{x^2} - 1} \right)\left( {{x^2} + 2x} \right)\\
= {x^4} + 2{x^3} - {x^2} - 2x\\
b)\left( {2x - 1} \right)\left( {3x + 2} \right)\left( {3 - x} \right)\\
= \left( {6{x^2} + 4x - 3x - 2} \right)\left( {3 - x} \right)\\
= \left( {6{x^2} + x - 2} \right)\left( {3 - x} \right)\\
= 18{x^2} - 6{x^3} + 3x - {x^2} - 6 + 2x\\
= - 6{x^3} + 17{x^2} + 5x - 6\\
c)\left( {x + 3} \right)\left( {{x^2} + 3x - 5} \right)\\
= {x^3} + 3{x^2} - 5x + 3{x^2} + 9x - 15\\
= {x^3} + 6{x^2} + 4x - 15\\
d)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\\
= {x^3} - {x^2} + x + {x^2} - x + 1\\
= {x^3} + 1\\
e)\left( {2{x^3} - 3x - 1} \right)\left( {5x + 2} \right)\\
= 10{x^4} + 4{x^3} - 15{x^2} - 6x - 5x - 2\\
= 10{x^4} + 4{x^4} - 15{x^2} - 11x - 2\\
f)\left( {{x^2} - 2x + 3} \right)\left( {x - 4} \right)\\
= {x^3} - 4{x^2} - 2{x^2} + 8x + 3x - 12\\
= {x^3} - 6{x^2} + 11x - 12
\end{array}$
