Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
16{x^4}\left( {x - y} \right) - x + y\\
 = 16{x^4}\left( {x - y} \right) - \left( {x - y} \right)\\
 = \left( {x - y} \right).\left( {16{x^4} - 1} \right)\\
 = \left( {x - y} \right).\left[ {{{\left( {4{x^2}} \right)}^2} - {1^2}} \right]\\
 = \left( {x - y} \right).\left( {4{x^2} - 1} \right)\left( {4{x^2} + 1} \right)\\
 = \left( {x - y} \right)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\\
b,\\
2{x^3}y - 2x{y^3} - 4x{y^2} - 2xy\\
 = 2xy.\left( {{x^2} - {y^2} - 2xy - 1} \right)\\
c,\\
x\left( {{y^2} - {z^2}} \right) + y\left( {{z^2} - {x^2}} \right) + z.\left( {{x^2} - {y^2}} \right)\\
 = x.\left( {y - z} \right).\left( {y + z} \right) + y{z^2} - y{x^2} + z{x^2} - z{y^2}\\
 = x.\left( {y - z} \right).\left( {y + z} \right) + \left( {y{z^2} - z{y^2}} \right) + \left( {z{x^2} - y{x^2}} \right)\\
 = x\left( {y - z} \right)\left( {y + z} \right) + yz\left( {z - y} \right) + {x^2}\left( {z - y} \right)\\
 = \left( {y - z} \right).\left[ {x\left( {y + z} \right) - yz - {x^2}} \right]\\
 = \left( {y - z} \right).\left( {xy + xz - yz - {x^2}} \right)\\
 = \left( {y - z} \right).\left[ {\left( {xy - yz} \right) + \left( {xz - {x^2}} \right)} \right]\\
 = \left( {y - z} \right).\left[ {y\left( {x - z} \right) + x.\left( {z - x} \right)} \right]\\
 = \left( {y - z} \right).\left( {x - z} \right).\left( {y - z} \right)\\
2,\\
a,\\
16{x^3} - 54{y^3} = 2.\left( {8{x^3} - 27{y^3}} \right)\\
 = 2.\left[ {{{\left( {2x} \right)}^3} - {{\left( {3y} \right)}^3}} \right]\\
 = 2.\left( {2x - 3y} \right)\left( {4{x^2} + 2x.3y + 9{y^2}} \right)\\
 = 2.\left( {2x - 3y} \right).\left( {4{x^2} + 6xy + 9{y^2}} \right)\\
b,\\
5{x^2} - 5{y^2} = 5.\left( {{x^2} - {y^2}} \right) = 5.\left( {x - y} \right)\left( {x + y} \right)\\
c,\\
16{x^3}y + y{z^3} = y.\left( {16{x^3} + {y^3}} \right)\\
d,\\
2{x^4} - 32 = 2.\left( {{x^4} - 16} \right) = 2.\left( {{x^4} - {4^2}} \right)\\
 = 2.\left( {{x^2} - 4} \right).\left( {{x^2} + 4} \right)\\
 = 2.\left( {x - 2} \right)\left( {x + 2} \right).\left( {{x^2} + 4} \right)
\end{array}\)