Đáp án:
$\begin{array}{l}
a){\left( {a - b - c} \right)^2} - {\left( {a - b + c} \right)^2}\\
= \left( {a - b - c + a - b + c} \right)\left( {a - b - c - a + b - c} \right)\\
= \left( {2a - 2b} \right).\left( { - 2c} \right)\\
= - 4ac + 4bc\\
b){\left( {a - x - y} \right)^3} - {\left( {a + x - y} \right)^3}\\
= {\left( {a - y - x} \right)^3} - {\left( {a - y + x} \right)^3}\\
= {\left( {a - y} \right)^3} - 3{\left( {a - y} \right)^2}.x + 3\left( {a - y} \right){x^2} - {x^3}\\
- \left( {{{\left( {a - y} \right)}^3} + 3{{\left( {a - y} \right)}^2}x + 3\left( {a - y} \right){x^2} + {x^3}} \right)\\
= - 6{\left( {a - y} \right)^2}x - 2{x^3}\\
= - 6{a^2}x + 12ayx - 6{y^2}x - 2{x^3}\\
c)\left( {1 + x + {x^2}} \right)\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)\\
= \left( {1 - {x^3}} \right)\left( {1 + {x^3}} \right)\\
= 1 - {x^6}
\end{array}$
