Đáp án:

$\begin{array}{l}
a)4{a^2}{b^2} - {\left( {{a^2} + {b^2} - 1} \right)^2}\\
 = {\left( {2ab} \right)^2} - {\left( {{a^2} + {b^2} - 1} \right)^2}\\
 = \left( {2ab - {a^2} - {b^2} + 1} \right)\left( {2ab + {a^2} + {b^2} - 1} \right)\\
 = \left( {1 - {{\left( {a - b} \right)}^2}} \right)\left( {{{\left( {a + b} \right)}^2} - 1} \right)\\
 = \left( {1 - a + b} \right)\left( {1 + a - b} \right)\left( {a + b + 1} \right)\left( {a + b - 1} \right)\\
b)\\
{\left( {{a^2} + {b^2} + ab} \right)^2} - {a^2}{b^2} - {b^2}{c^2} - {c^2}{a^2}\\
 = \left( {{a^2} + {b^2} + ab - ab} \right)\left( {{a^2} + {b^2} + ab + ab} \right)\\
 - {c^2}\left( {{a^2} + {b^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2}} \right) - {c^2}\left( {{a^2} + {b^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2} - {c^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left[ {{{\left( {a + b} \right)}^2} - {c^2}} \right]\\
 = \left( {{a^2} + {b^2}} \right)\left( {a + b + c} \right)\left( {a + b - c} \right)
\end{array}$

$\begin{array}{l} a)4{a^2}{b^2} - {\left( {{a^2} + {b^2} - 1} \right)^2}\\  = {\left( {2ab} \right)^2} - {\left( {{a^2} + {b^2} - 1} \right)^2}\\  = \left( {2ab - {a^2} - {b^2} + 1} \right)\left( {2ab + {a^2} + {b^2} - 1} \right)\\  = \left( {1 - a + b} \right)\left( {1 + a - b} \right)\left( {a + b + 1} \right)\left( {a + b - 1} \right)\\ b)\\ {\left( {{a^2} + {b^2} + ab} \right)^2} - {a^2}{b^2} - {b^2}{c^2} - {c^2}{a^2}\\  = \left( {{a^2} + {b^2} + ab - ab} \right)\left( {{a^2} + {b^2} + ab + ab} \right)\\  - {c^2}\left( {{a^2} + {b^2}} \right)\\  = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2}} \right) - {c^2}\left( {{a^2} + {b^2}} \right)\\  = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2} - {c^2}} \right)\\  = \left( {{a^2} + {b^2}} \right)\left( {a + b + c} \right)\left( {a + b - c} \right) \end{array}$

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