Đáp án:
$\begin{array}{l}
d){\left( {{x^2} + {y^2} + xy} \right)^2} - {x^2}{y^2} - {y^2}{z^2} - {x^2}{z^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} + 2xy\left( {{x^2} + {y^2}} \right) + {x^2}{y^2}\\
- {x^2}{y^2} - {y^2}{z^2} - {x^2}{z^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} + 2xy\left( {{x^2} + {y^2}} \right) - {z^2}\left( {{x^2} + {y^2}} \right)\\
= \left( {{x^2} + {y^2}} \right)\left( {{x^2} + {y^2} + 2xy - {z^2}} \right)\\
= \left( {{x^2} + {y^2}} \right)\left[ {{{\left( {x + y} \right)}^2} - {z^2}} \right]\\
= \left( {{x^2} + {y^2}} \right)\left( {x + y + z} \right)\left( {x + y - z} \right)\\
e)3x - 3y + {x^2} - 2xy + {y^2}\\
= 3\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
= \left( {x - y} \right)\left( {3 + x - y} \right)\\
f){x^2} + 2xy + {y^2} - 2x - 2y + 1\\
= {\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 1\\
= {\left( {x + y - 1} \right)^2}
\end{array}$
