Đáp án:
$\begin{array}{l}
a){\left( {a - b} \right)^3} = {\left[ { - \left( {b - a} \right)} \right]^3}\\
= {\left( { - 1} \right)^3}.{\left( {b - a} \right)^3}\\
= - {\left( {b - a} \right)^3}\\
b){\left( { - a - b} \right)^2} = {\left[ { - \left( {a + b} \right)} \right]^2}\\
= {\left( { - 1} \right)^2}.{\left( {a + b} \right)^2}\\
= {\left( {a + b} \right)^2}\\
c){\left( {x + y} \right)^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
x{\left( {x - 3y} \right)^2} + y{\left( {y - 3x} \right)^2}\\
= x\left( {{x^2} - 6xy + 9{y^2}} \right) + y\left( {{y^2} - 6xy + 9{x^2}} \right)\\
= {x^3} - 6{x^2}y + 9x{y^2} + {y^3} - 6x{y^2} + 9{x^2}y\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
\Leftrightarrow {\left( {x + y} \right)^3} = x{\left( {x - 3y} \right)^2} + y{\left( {y - 3x} \right)^2}\\
d){\left( {x + y} \right)^3} - {\left( {x - y} \right)^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3} - {x^3} + 3{x^2}y - 3x{y^2} + {y^3}\\
= 6{x^2}y + 2{y^3}\\
= 2y\left( {3{x^2} + {y^2}} \right)\\
e){a^3} + {b^3}\\
{\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\\
= {a^3} + 3{a^2}b + 3a{b^2} + {b^3} - 3{a^2}b - 3a{b^2}\\
= {a^3} + {b^3}\\
f){\left( {a - b} \right)^3} - 3ab\left( {a - b} \right)\\
= {a^3} - 3{a^2}b + 3a{b^2} - {b^3} - 3{a^2}b + 3a{b^2}\\
= {a^3} - {b^3}
\end{array}$
