Đáp án:
$\begin{array}{l}
a)\left( {a + b + c + d} \right)\left( {a - b - c - d} \right)\\
= {a^2} - {\left( {b + c + d} \right)^2}\\
= {a^2} - \left( {{b^2} + {c^2} + {d^2} + 2bc + 2cd + 2bd} \right)\\
= {a^2} - {b^2} - {c^2} - {d^2} - 2bc - 2cd - 2bd\\
b)\left( {a - b + c + d} \right)\left( {a - b - c - d} \right)\\
= {\left( {a - b} \right)^2} - {\left( {c + d} \right)^2}\\
= {a^2} - 2ab + {b^2} - {c^2} - 2cd - {d^2}\\
c)\left( {a - b - c + d} \right)\left( {a + b - c - d} \right)\\
= {\left( {a - c} \right)^2} - {\left( {b - d} \right)^2}\\
= {a^2} - 2ac + {c^2} - {b^2} + 2bd - {d^2}\\
d)\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab - ac - bc} \right)\\
= {a^3} + a{b^2} + a{c^2} + 2{a^2}b - {a^2}c - abc\\
+ {a^2}b + {b^3} + b{c^2} + 2a{b^2} - abc - {b^2}c\\
+ {a^2}c + {b^2}c + {c^3} + 2abc - a{c^2} - b{c^2}\\
= {a^3} + {b^3} + {c^3} + 3a{b^2} + 3{a^2}b\\
e)\left( {a - b - c} \right)\left( {{a^2} + {b^2} + {c^2} - 2ab + ac - bc} \right)\\
= {a^3} + a{b^2} + a{c^2} - 2{a^2}b + {a^2}c - abc\\
- {a^2}b - {b^3} - b{c^2} + 2a{b^2} - abc + {b^2}c\\
- {a^2}c - {b^2}c - {c^3} + 2abc - a{c^2} + b{c^2}\\
= {a^3} - {b^3} - {c^3} + 3a{b^2} - 3{a^2}b
\end{array}$
