Đáp án:
$\begin{array}{l}
VD2:\\
a){x^6} - 3{x^4}y + 3{x^2}{y^2} - {y^3}\\
= {\left( {{x^2} - y} \right)^3}\\
b){\left( {x - y} \right)^3} + {\left( {x - y} \right)^2} + \dfrac{1}{3}\left( {x - y} \right) + \dfrac{1}{{27}}\\
= {\left( {x - y + \dfrac{1}{3}} \right)^3}\\
VD3)\\
a)B = {\left( {xy + 2} \right)^3} - 6{\left( {xy + 2} \right)^2} + 12\left( {xy + 2} \right) - 8\\
= {\left( {xy + 2 - 2} \right)^3}\\
= {\left( {xy} \right)^3}\\
= {x^3}{y^3}\\
VD4)a)A = 8{x^3} + 12{x^2} + 6x + 1\\
= {\left( {2x + 1} \right)^3}\\
= {\left( {2.\dfrac{1}{2} + 1} \right)^3}\\
= {2^3}\\
= 8\\
b)C = {\left( {x + 2y} \right)^3} - 6{\left( {x + 2y} \right)^2} + 12\left( {x + 2y} \right) - 8\\
= {\left( {x + 2y - 2} \right)^3}\\
= {\left( {20 + 2.1 - 2} \right)^3}\\
= {20^3}\\
= 8000
\end{array}$
