Đáp án:
$\begin{array}{l}
a){x^3} + 64\\
= {x^3} + {4^3}\\
= \left( {x + 4} \right)\left( {{x^2} - x.4 + {4^2}} \right)\\
= \left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right)\\
b){\left( {\dfrac{1}{2}x - 2{y^2}} \right)^3}\\
= {\left( {\dfrac{1}{2}x} \right)^3} + 3.{\left( {\dfrac{1}{2}x} \right)^2}.\left( { - 2{y^2}} \right)\\
+ 3.\left( {\dfrac{1}{2}x} \right).{\left( { - 2y} \right)^2} + {\left( { - 2y} \right)^3}\\
= \dfrac{1}{8}{x^3} - \dfrac{3}{2}{x^2}{y^2} + 6x{y^2} - 8{y^3}\\
c)\left( {4{x^2} + \dfrac{1}{3}} \right).\left( {16{x^4} - \dfrac{4}{3}{x^2} + \dfrac{1}{9}} \right)\\
= \left( {4{x^2} + \dfrac{1}{3}} \right).\left[ {{{\left( {4{x^2}} \right)}^2} - 4{x^2}.\dfrac{1}{3} + {{\left( {\dfrac{1}{3}} \right)}^2}} \right]\\
= {\left( {4{x^2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3}\\
= 64{x^6} + \dfrac{1}{{27}}
\end{array}$
