Làm từng bước đầy đủ giúp mình

Đáp án:

$\begin{array}{l}
1)a){x^3} + 8\\
 = {x^3} + {2^3}\\
 = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
b){x^3} - 64\\
 = {x^3} - {4^3}\\
 = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right)\\
c)8{x^3} + 1\\
 = {\left( {2x} \right)^3} + 1\\
 = \left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right)\\
d)27 - {x^3}\\
 = \left( {3 - x} \right)\left( {9 + 3x + {x^2}} \right)\\
e)125 + 8{x^3}\\
 = \left( {5 + 2x} \right)\left( {25 - 10x + 4{x^2}} \right)\\
f){x^9} - 27{y^3}\\
 = {\left( {{x^3}} \right)^3} - {\left( {3y} \right)^3}\\
 = \left( {{x^3} - 3y} \right)\left( {{x^6} + 3{x^3}y + 9{y^2}} \right)\\
2)a)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
 = {x^3} + {2^3}\\
 = {x^3} + 8\\
b)\left( {2 - x} \right)\left( {{x^2} + 2x + 4} \right)\\
 = {2^3} - {x^3}\\
 = 8 - {x^3}\\
c)\left( {x + 3y} \right)\left( {9{y^2} - 3xy + {x^2}} \right)\\
 = {x^3} + {\left( {3y} \right)^3}\\
 = {x^3} + 27{y^3}\\
d)\left( {4 - \dfrac{x}{2}} \right)\left( {\dfrac{{{x^2}}}{4} + 2x + 16} \right)\\
 = {4^3} - {\left( {\dfrac{x}{2}} \right)^3}\\
 = 64 - \dfrac{{{x^3}}}{8}\\
e)\left( {x + \dfrac{1}{3}} \right)\left( {{x^2} - \dfrac{x}{3} + \dfrac{1}{9}} \right)\\
 = {x^3} + {\left( {\dfrac{1}{3}} \right)^3}\\
 = {x^3} + \dfrac{1}{{27}}\\
f)\left( {\dfrac{1}{4} - \dfrac{x}{5}} \right)\left( {\dfrac{{{x^2}}}{{25}} + \dfrac{x}{{20}} + \dfrac{1}{{16}}} \right)\\
 = {\left( {\dfrac{1}{4}} \right)^3} - {\left( {\dfrac{x}{5}} \right)^3}\\
 = \dfrac{1}{{64}} - \dfrac{{{x^3}}}{{125}}
\end{array}$