Đáp án:
\[\dfrac{8}{{27}}{x^6} - \dfrac{2}{3}{x^4}y + \dfrac{1}{2}{x^2}{y^2} - \dfrac{1}{8}{y^3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {\dfrac{2}{3}{x^2} - \dfrac{1}{2}y} \right)^3}\\
= {\left( {\dfrac{2}{3}{x^2}} \right)^3} - 3.{\left( {\dfrac{2}{3}{x^2}} \right)^2}.\dfrac{1}{2}y + 3.\dfrac{2}{3}{x^2}.{\left( {\dfrac{1}{2}y} \right)^2} - {\left( {\dfrac{1}{2}y} \right)^3}\\
= {\left( {\dfrac{2}{3}} \right)^3}.{\left( {{x^2}} \right)^3} - 3.\dfrac{4}{9}{x^4}.\dfrac{1}{2}y + 3.\dfrac{2}{3}{x^2}.\dfrac{1}{4}{y^2} - \dfrac{1}{8}{y^3}\\
= \dfrac{8}{{27}}{x^6} - \dfrac{2}{3}{x^4}y + \dfrac{1}{2}{x^2}{y^2} - \dfrac{1}{8}{y^3}
\end{array}\)
