Đáp án:

$\begin{array}{l}
a){x^2}y + x{y^2} - x - y\\
 = xy\left( {x + y} \right) - \left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {xy - 1} \right)\\
b)a.{x^2} - a.{y^2} - 7x - 7y\\
 = a.\left( {{x^2} - {y^2}} \right) - 7.\left( {x + y} \right)\\
 = \left( {x + y} \right).\left( {a.\left( {x - y} \right) - 7} \right)\\
 = \left( {x + y} \right)\left( {a.x - a.y - 7} \right)\\
c)a.{x^2} + a.y - b{x^2} - by\\
 = {x^2}\left( {a - b} \right) + y\left( {a - b} \right)\\
 = \left( {a - b} \right)\left( {{x^2} + y} \right)\\
d)x{\left( {x + 1} \right)^2} + x\left( {x - 5} \right) - 5{\left( {x + 1} \right)^2}\\
 = {\left( {x + 1} \right)^2}\left( {x - 5} \right) + x\left( {x - 5} \right)\\
 = \left( {x - 5} \right)\left( {{x^2} + 2x + 1 + x} \right)\\
 = \left( {x - 5} \right)\left( {{x^2} + 3x + 1} \right)\\
e)ab\left( {a - b} \right) + bc\left( {b - c} \right) + ca\left( {c - a} \right)\\
 = {a^2}b - a{b^2} + {b^2}c - b{c^2} + ca\left( {c - a} \right)\\
 = b.\left( {{a^2} - {c^2}} \right) - {b^2}\left( {a - c} \right) - ca\left( {a - c} \right)\\
 = \left( {a - c} \right)\left( {b\left( {a + c} \right) - {b^2} - ca} \right)\\
 = \left( {a - c} \right)\left( {ba + bc - {b^2} - ca} \right)\\
 = \left( {a - c} \right)\left( {a - b} \right).\left( {b - c} \right)\\
g){\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3}\\
 = \left( {a + b + c - a} \right)\left[ {{{\left( {a + b + c} \right)}^2} + \left( {a + b + c} \right).a + {a^2}} \right]\\
 - \left( {b + c} \right)\left( {{b^2} - bc + {c^2}} \right)\\
 = \left( {b + c} \right).\left( \begin{array}{l}
{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac + {a^2} + ab + ac + {a^2}\\
 - {b^2} + bc - {c^2}
\end{array} \right)\\
 = \left( {b + c} \right)\left( {3{a^2} + 3ab + 3ac + 3bc} \right)
\end{array}$