Đáp án:

$\begin{array}{l}
a){x^2} - xy + {x^3} - 3{x^2}y + 3x{y^2} - {y^3}\\
 = x\left( {x - y} \right) + {\left( {x - y} \right)^3}\\
 = \left( {x - y} \right)\left( {x + {{\left( {x - y} \right)}^2}} \right)\\
 = \left( {x - y} \right) + ({x^2} + {y^2} + x - 2xy)\\
b)3{x^3} - {x^2} - 21x + 7\\
 = {x^2}\left( {3x - 1} \right) - 7\left( {3x - 1} \right)\\
 = \left( {3x - 1} \right)\left( {{x^2} - 7} \right)\\
c){x^3} - 4{x^2} + 8x - 8\\
 = {x^3} - 4{x^2} + 4x + 4x - 8\\
 = x\left( {{x^2} - 4x + 4} \right) + 4\left( {x - 2} \right)\\
 = x{\left( {x - 2} \right)^2} + 4\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x\left( {x - 2} \right) + 4} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} - 2x + 4} \right)\\
d){x^3} - 5{x^2} - 5x + 1\\
 = {x^3} + 1 - 5{x^2} - 5x\\
 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - 5x\left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - x + 1 - 5x} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - 6x + 1} \right)\\
e){x^2}y - xz + z - y\\
 = {x^2}y - y + z - xz\\
 = y\left( {{x^2} - 1} \right) - z\left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {y.\left( {x + 1} \right) - z} \right)\\
 = \left( {x - 1} \right)\left( {xy + y - z} \right)\\
f){x^4} - {x^3} + {x^2} - 1\\
 = {x^3}\left( {x - 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^3} + x + 1} \right)\\
g){x^4} - {x^2} + 10x - 25\\
 = {x^4} - \left( {{x^2} - 10x + 25} \right)\\
 = {x^4} - {\left( {x - 5} \right)^2}\\
 = \left( {{x^2} - x + 5} \right)\left( {{x^2} + x - 5} \right)
\end{array}$