Đáp án:
$\begin{array}{l}
1){x^3} + 8 = {x^3} + {2^3} = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
2)27 - {z^3} = \left( {3 - z} \right)\left( {9 + 3z + {z^2}} \right)\\
3)8{x^3} - 64\\
= 8.\left( {{x^3} - 8} \right)\\
= 8\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\\
4){x^6} - {y^3}\\
= \left( {{x^2} - y} \right)\left( {{x^4} + {x^2}y + {y^2}} \right)\\
5){x^6} - {y^6}\\
= \left( {{x^2} - {y^2}} \right)\left( {{x^4} + {x^2}{y^2} + {y^4}} \right)\\
= \left( {x - y} \right)\left( {x + y} \right)\left( {{x^4} + {x^2}{y^2} + {y^4}} \right)\\
6)1 + 8{x^6}{y^3}\\
= \left( {1 + 2{x^2}y} \right)\left( {1 - 2{x^2}y + 4{x^4}{y^2}} \right)\\
7)125{x^3} + 1\\
= \left( {5x + 1} \right)\left( {25{x^2} - 5x + 1} \right)\\
8)8{x^3} - 27\\
= \left( {2x - 3} \right)\left( {4{x^2} + 6x + 9} \right)\\
9)125{x^3} + 27{y^3}\\
= {\left( {5x} \right)^3} + {\left( {3y} \right)^3}\\
= \left( {5x + 3y} \right)\left( {25{x^2} - 15xy + 9{y^2}} \right)\\
10){x^3} + \dfrac{1}{{27}}\\
= \left( {x + \dfrac{1}{3}} \right)\left( {{x^2} - \dfrac{1}{3}x + \dfrac{1}{9}} \right)\\
11)27{x^3} + \dfrac{{{y^3}}}{8}\\
= \left( {3x + \dfrac{y}{2}} \right)\left( {9{x^2} - \dfrac{{3xy}}{2} + \dfrac{{{y^2}}}{4}} \right)\\
13){\left( {a + b} \right)^3} - {\left( {a - b} \right)^3}\\
= \left( {a + b - a + b} \right)\left[ {{{\left( {a + b} \right)}^2} + \left( {a + b} \right)\left( {a - b} \right) + {{\left( {a - b} \right)}^3}} \right]\\
= 2b.\left( {{a^2} + 2ab + {b^2} + {a^2} - {b^2} + {a^2} - 2ab + {b^2}} \right)\\
= 2b.\left( {3{a^2} + {b^2}} \right)
\end{array}$
