Đáp án:
$\begin{array}{l}
a){\left( {{x^2} + 4{y^2} - 20} \right)^2} - 16{\left( {xy - 4} \right)^2}\\
= {\left( {{x^2} + 4{y^2} - 20} \right)^2} - {\left( {4xy - 16} \right)^2}\\
= \left( {{x^2} + 4{y^2} - 20 - 4xy + 16} \right)\\
\left( {{x^2} + 4{y^2} - 20 + 4xy - 16} \right)\\
= \left[ {{{\left( {x - 2y} \right)}^2} - 4} \right]\left[ {{{\left( {x + 2y} \right)}^2} - 36} \right]\\
= \left( {x - 2y + 4} \right)\left( {x - 2y - 4} \right)\\
.\left( {x + 2y - 6} \right)\left( {x + 2y + 6} \right)\\
b){\left( {x + y} \right)^3} - 1 - 3\left( {x + y} \right)\left( {x + y - 1} \right)\\
= \left( {x + y - 1} \right)\left[ {{{\left( {x + y} \right)}^2} + x + y + 1} \right]\\
- \left( {3x + 3y} \right)\left( {x + y - 1} \right)\\
= \left( {x + y - 1} \right)\\
.\left[ {{x^2} + 2xy + {y^2} + x + y + 1 - 3x - 3y} \right]\\
= \left( {x + y - 1} \right).\left( {{x^2} + {y^2} - 2x - 2y - 2xy + 1} \right)
\end{array}$
