Đáp án:
$\begin{array}{l}
1){x^6} - {x^4} + 2{x^3} + 2{x^2}\\
= {x^2}\left( {{x^4} - {x^2} + 2x + 1} \right)\\
2){x^3}{\left( {x - 3} \right)^2} - {\left( {x - 3} \right)^2} - {x^2} + 1\\
= {\left( {x - 3} \right)^2}\left( {{x^3} - 1} \right) - \left( {{x^2} - 1} \right)\\
= {\left( {x - 3} \right)^2}.\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right)\\
= \left( {x - 1} \right).\left[ {{{\left( {x - 3} \right)}^2}.\left( {{x^2} + x + 1} \right) - x - 1} \right]\\
= \left( {x - 1} \right).\left[ {\left( {{x^2} - 6x + 9} \right)\left( {{x^2} + x + 1} \right) - x - 1} \right]\\
= \left( {x - 1} \right)\left( {{x^4} - 5{x^3} + 4{x^2} + 3x + 9 - x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^4} - 5{x^3} + 4{x^2} + 2x + 8} \right)\\
3){x^3} - 2{x^2} + 4x - 8\\
= {x^2}\left( {x - 2} \right) + 4\left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {{x^2} + 4} \right)\\
5){\left( {x + y} \right)^3} - {\left( {x - y} \right)^3}\\
= \left( {x + y - x + y} \right)\left[ {{{\left( {x + y} \right)}^2} + \left( {x + y} \right)\left( {x - y} \right) + {{\left( {x - y} \right)}^2}} \right]\\
= 2y.\left( {{x^2} + 2xy + {y^2} + {x^2} - {y^2} + {x^2} - 2xy + {y^2}} \right)\\
= 2y.\left( {3{x^2} + {y^2}} \right)
\end{array}$
