Đáp án:

\(\begin{array}{l}
e,\\
64 - 96b + 48{b^2} - 8{b^3}\\
f,\\
8{c^2} - 36{c^2}d + 54c{d^2} - 27{d^3}\\
g,\\
\dfrac{{27{x^3}}}{{{y^3}}} - \dfrac{{54x}}{y} + \dfrac{{36y}}{x} - \dfrac{{8{y^3}}}{{{x^3}}}\\
h,\\
{x^3} + \dfrac{3}{5}{x^2} + \dfrac{3}{{25}}x + \dfrac{1}{{125}}\\
i,\\
\left( {3 - y} \right)\left( {9 + 3y + {y^2}} \right)\\
k,\\
\left( {5 + t} \right)\left( {25 - 10t + {t^2}} \right)\\
m,\\
\left( {4m - 3} \right)\left( {16{m^2} - 36m + 9} \right)\\
n,\\
\left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
e,\\
{\left( {4 - 2b} \right)^3}\\
 = {4^3} - {3.4^2}.2b + 3.4.{\left( {2b} \right)^2} - {\left( {2b} \right)^3}\\
 = 64 - 3.16.2b + 3.4.4{b^2} - 8{b^3}\\
 = 64 - 96b + 48{b^2} - 8{b^3}\\
f,\\
{\left( {2c - 3d} \right)^3}\\
 = {\left( {2c} \right)^3} - 3.{\left( {2c} \right)^2}.3d + 3.2c.{\left( {3d} \right)^2} - {\left( {3d} \right)^3}\\
 = 8{c^2} - 3.4{c^2}.3d + 3.2c.9{d^2} - 27{d^3}\\
 = 8{c^2} - 36{c^2}d + 54c{d^2} - 27{d^3}\\
g,\\
{\left( {\dfrac{{3x}}{y} - \dfrac{{2y}}{x}} \right)^3}\\
 = {\left( {\dfrac{{3x}}{y}} \right)^3} - 3.{\left( {\dfrac{{3x}}{y}} \right)^2}.\dfrac{{2y}}{x} + 3.\dfrac{{3x}}{y}.{\left( {\dfrac{{2y}}{x}} \right)^2} - {\left( {\dfrac{{2y}}{x}} \right)^3}\\
 = \dfrac{{27{x^3}}}{{{y^3}}} - 3.\dfrac{{9{x^2}}}{{{y^2}}}.\dfrac{{2y}}{x} + 3.\dfrac{{3x}}{y}.\dfrac{{4{y^2}}}{{{x^2}}} - \dfrac{{8{y^3}}}{{{x^3}}}\\
 = \dfrac{{27{x^3}}}{{{y^3}}} - \dfrac{{54x}}{y} + \dfrac{{36y}}{x} - \dfrac{{8{y^3}}}{{{x^3}}}\\
h,\\
{\left( {x + \dfrac{1}{5}} \right)^3}\\
 = {x^3} + 3.{x^2}.\dfrac{1}{5} + 3.x.{\left( {\dfrac{1}{5}} \right)^2} + {\left( {\dfrac{1}{5}} \right)^3}\\
 = {x^3} + \dfrac{3}{5}{x^2} + \dfrac{3}{{25}}x + \dfrac{1}{{125}}\\
i,\\
27 - {y^3}\\
 = {3^2} - {y^3}\\
 = \left( {3 - y} \right).\left( {{3^2} + 3.y + {y^2}} \right)\\
 = \left( {3 - y} \right)\left( {9 + 3y + {y^2}} \right)\\
k,\\
125 + {t^3}\\
 = {5^3} + {t^3}\\
 = \left( {5 + t} \right).\left( {{5^2} - 5.t + {t^2}} \right)\\
 = \left( {5 + t} \right)\left( {25 - 10t + {t^2}} \right)\\
m,\\
64{m^3} - 27\\
 = {\left( {4m} \right)^3} - {3^3}\\
 = \left( {4m - 3} \right).\left[ {{{\left( {4m} \right)}^2} - 3.4m.3 + {3^2}} \right]\\
 = \left( {4m - 3} \right)\left( {16{m^2} - 36m + 9} \right)\\
n,\\
8{x^3} + 27\\
 = {\left( {2x} \right)^3} + {3^3}\\
 = \left( {2x + 3} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.3 + {3^2}} \right]\\
 = \left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right)
\end{array}\)