Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
6,\\
a,\\
{\left( {3x + 1} \right)^3}\\
 = {\left( {3x} \right)^3} + 3.{\left( {3x} \right)^2}.1 + 3.3x{.1^2} + {1^3}\\
 = 27{x^3} + 37{x^2} + 9x + 1\\
b,\\
{\left( {2x - \dfrac{1}{x}} \right)^3}\\
 = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.\dfrac{1}{x} + 3.2x.{\left( {\dfrac{1}{x}} \right)^2} - {\left( {\dfrac{1}{x}} \right)^3}\\
 = 8{x^3} - 3.4{x^2}.\dfrac{1}{x} + 6x.\dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}\\
 = 8{x^3} - 12x + \dfrac{6}{x} - \dfrac{1}{{{x^3}}}\\
c,\\
{\left( {y - \dfrac{{xy}}{3}} \right)^3}\\
 = {y^3} - 3{y^2}.\dfrac{{xy}}{3} + 3y.{\left( {\dfrac{{xy}}{3}} \right)^2} - {\left( {\dfrac{{xy}}{3}} \right)^3}\\
 = {y^3} - x{y^3} + 3y.\dfrac{{{x^2}{y^2}}}{9} - \dfrac{{{x^3}{y^3}}}{{27}}\\
 = {y^3} - x{y^3} + \dfrac{1}{3}{x^2}{y^3} - \dfrac{1}{{27}}{x^3}{y^3}\\
d,\\
{\left( {\dfrac{1}{{{y^2}}} + \dfrac{y}{x}} \right)^3}\\
 = {\left( {\dfrac{1}{{{y^2}}}} \right)^3} + 3.{\left( {\dfrac{1}{{{y^2}}}} \right)^2}.\dfrac{y}{x} + 3.\dfrac{1}{{{y^2}}}.{\left( {\dfrac{y}{x}} \right)^2} + {\left( {\dfrac{y}{x}} \right)^3}\\
 = \dfrac{1}{{{y^6}}} + 3.\dfrac{1}{{{y^4}}}.\dfrac{y}{x} + 3.\dfrac{1}{{{y^2}}}.\dfrac{{{y^2}}}{{{x^2}}} + \dfrac{{{y^3}}}{{{x^3}}}\\
 = \dfrac{1}{{{y^6}}} + \dfrac{3}{{x{y^3}}} + \dfrac{3}{{{x^2}}} + \dfrac{{{y^3}}}{{{x^3}}}\\
7,\\
a,\\
A = 8{x^6} - 12{x^4} + 6{x^2} - 1\\
 = {\left( {2{x^2}} \right)^3} - 3.{\left( {2{x^2}} \right)^2}.1 + 3.2{x^2}{.1^2} - {1^3}\\
 = {\left( {2{x^2} - 1} \right)^3}\\
b,\\
B = 8{\left( {\dfrac{x}{2} + y} \right)^3} - 6.{\left( {x + 2y} \right)^2}.x + 12.\left( {x + 2y} \right).{x^2} - 8{x^3}\\
 = {\left[ {2.\left( {\dfrac{x}{y} + y} \right)} \right]^3} - 3.{\left( {x + 2y} \right)^2}.2x + 3.\left( {x + 2y} \right).4{x^2} - {\left( {2x} \right)^3}\\
 = {\left( {x + 2y} \right)^3} - 3.{\left( {x + 2y} \right)^2}.2x + 3.\left( {x + 2y} \right).{\left( {2x} \right)^2} - {\left( {2x} \right)^3}\\
 = {\left[ {\left( {x + 2y} \right) - 2x} \right]^3}\\
 = {\left( {2y - x} \right)^3}\\
c,\\
C = {\left( {x - y} \right)^3} - \dfrac{{3{{\left( {x - y} \right)}^2}}}{2}y + \dfrac{{3\left( {x - y} \right)}}{4}{y^2} - \dfrac{{{y^3}}}{8}\\
 = {\left( {x - y} \right)^3} - 3.{\left( {x - y} \right)^2}.\dfrac{y}{2} + 3.\left( {x - y} \right).{\left( {\dfrac{y}{2}} \right)^2} - {\left( {\dfrac{y}{2}} \right)^3}\\
 = {\left[ {\left( {x - y} \right) - \dfrac{y}{2}} \right]^3}\\
 = {\left( {x - \dfrac{3}{2}y} \right)^3}
\end{array}\)