Đáp án:
$\begin{array}{l}
1)\left( {8{a^3} - 27{b^3}} \right) - 2a\left( {4{a^2} - 2{b^2}} \right)\\
= 8{a^3} - 27{b^3} - 8{a^3} + 4a{b^2}\\
= 4a{b^2} - 27{b^3}\\
= {b^2}\left( {4a - 27b} \right)\\
2){\left( {3x - 1} \right)^2} - 16\\
= {\left( {3x - 1} \right)^2} - {4^2}\\
= \left( {3x - 1 - 4} \right)\left( {3x - 1 + 4} \right)\\
= \left( {3x - 5} \right)\left( {3x + 3} \right)\\
3){\left( {3x + 1} \right)^2} - 4{\left( {x - 2} \right)^2}\\
= \left( {3x + 1 - 2\left( {x - 2} \right)} \right)\left( {3x + 1 + 2\left( {x - 2} \right)} \right)\\
= \left( {x + 5} \right)\left( {5x - 3} \right)\\
4)1 - {\left( {{x^2} - 2xy + {y^2}} \right)^2}\\
= \left( {1 - {x^2} + 2xy - {y^2}} \right)\left( {1 + {x^2} - 2xy + {y^2}} \right)\\
= \left( {{x^2} - 2xy + {y^2} + 1} \right)\left( {1 - {{\left( {x - y} \right)}^2}} \right)\\
= \left( {{x^2} - 2xy + {y^2} + 1} \right)\left( {1 - x + y} \right)\left( {1 + x - y} \right)
\end{array}$
