Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^3} + 2{x^2}y + x{y^2} - 4x{z^2}\\
= x.\left( {{x^2} + 2xy + {y^2} - 4{z^2}} \right)\\
= x.\left[ {\left( {{x^2} + 2xy + {y^2}} \right) - 4{z^2}} \right]\\
= x.\left[ {{{\left( {x + y} \right)}^2} - {{\left( {2z} \right)}^2}} \right]\\
= x.\left( {x + y - 2z} \right).\left( {x + y + 2z} \right)\\
b,\\
- 8{x^3} + 12{x^2}y - 6x{y^2} + {y^3}\\
= {\left( { - 2x} \right)^3} + 3.{\left( { - 2x} \right)^2}.y + 2.\left( { - 2x} \right).{y^2} + {y^3}\\
= {\left( { - 2x + y} \right)^3}\\
c,\\
6{x^2} + 7x - 5\\
= \left( {6{x^2} - 3x} \right) + \left( {10x - 5} \right)\\
= 3x.\left( {2x - 1} \right) + 5.\left( {2x - 1} \right)\\
= \left( {2x - 1} \right)\left( {3x + 5} \right)\\
d,\\
{x^4} + 64{y^4}\\
= \left( {{x^4} + 16{x^2}{y^2} + 64{y^4}} \right) - 16{x^2}{y^2}\\
= \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.8{y^2} + {{\left( {8{y^2}} \right)}^2}} \right] - {\left( {4xy} \right)^2}\\
= \left( {{x^2} + 8{y^2}} \right) - {\left( {4xy} \right)^2}\\
= \left( {{x^2} + 8{y^2} - 4xy} \right)\left( {{x^2} + 8{y^2} + 4xy} \right)
\end{array}\)
