Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {\left( {x - y} \right)^3} + 3xy\left( {x - y} \right)\\
= \left( {{x^3} - 3{x^2}y + 3x{y^2} - {y^3}} \right) + \left( {3{x^2}y - 3x{y^2}} \right)\\
= {x^3} - 3{x^2}y + 3x{y^2} - {y^3} + 3{x^2}y - 3x{y^2}\\
= {x^3} - {y^3}\\
b,\\
B = {\left( {x + y} \right)^3} + 3.\left( {x - y} \right){\left( {x + y} \right)^2} + 3{\left( {x - y} \right)^2}.\left( {x + y} \right) + {\left( {x - y} \right)^3}\\
= {\left( {x + y} \right)^3} + 3.{\left( {x + y} \right)^2}.\left( {x - y} \right) + 3.\left( {x + y} \right).{\left( {x - y} \right)^2} + {\left( {x - y} \right)^3}\\
= {\left[ {\left( {x + y} \right) + \left( {x - y} \right)} \right]^3}\\
= {\left( {2x} \right)^3}\\
= 8{x^3}
\end{array}\)
