Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} - 16 - 4xy + 4{y^2}\\
= \left( {{x^2} - 4xy + 4{y^2}} \right) - 16\\
= {\left( {x - 2y} \right)^2} - {4^2}\\
= \left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
b,\\
{x^2} - 3{x^3} - x + 3\\
= \left( { - 3{x^3} + 3} \right) + \left( {{x^2} - x} \right)\\
= - 3.\left( {{x^3} - 1} \right) + x\left( {x - 1} \right)\\
= - 3.\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + x\left( {x - 1} \right)\\
= \left( {x - 1} \right).\left[ { - 3\left( {{x^2} + x + 1} \right) + x} \right]\\
= \left( {x - 1} \right)\left( { - 3{x^2} - 2x - 3} \right)\\
c,\\
{x^3} - {x^2}y - x{y^2} + {y^3}\\
= \left( {{x^3} - {x^2}y} \right) - \left( {x{y^2} - {y^3}} \right)\\
= {x^2}\left( {x - y} \right) - {y^2}\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {{x^2} - {y^2}} \right)\\
= \left( {x - y} \right)\left( {x - y} \right)\left( {x + y} \right)\\
= {\left( {x - y} \right)^2}\left( {x + y} \right)\\
d,\\
3x + 3y - {x^2} - 2xy - {y^2}\\
= \left( {3x + 3y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
= 3.\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
= \left( {x + y} \right)\left( {3 - x - y} \right)\\
e,\\
4{x^4} + 4{x^3} - {x^2} - x\\
= \left( {4{x^4} + 4{x^3}} \right) - \left( {{x^2} + x} \right)\\
= 4{x^3}\left( {x + 1} \right) - x\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {4{x^3} - x} \right)\\
= x.\left( {x + 1} \right)\left( {4{x^2} - 1} \right)\\
= x\left( {x + 1} \right)\left( {2x - 1} \right)\left( {2x + 1} \right)\\
f,\\
{x^4} - 4{x^3} + 8{x^2} - 16x + 16\\
= \left( {{x^4} - 4{x^3} + 4{x^2}} \right) + \left( {4{x^2} - 16x + 16} \right)\\
= {x^2}\left( {{x^2} - 4x + 4} \right) + 4\left( {{x^2} - 4x + 4} \right)\\
= \left( {{x^2} - 4x + 4} \right)\left( {{x^2} + 4} \right)\\
= {\left( {x - 2} \right)^2}.\left( {{x^2} + 4} \right)\\
g,\\
{a^3} + {b^3} + {c^3} - 3abc\\
= \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) + {c^3} - \left( {3{a^2}b + 3a{b^2} + 3abc} \right)\\
= {\left( {a + b} \right)^3} + {c^3} - 3ab\left( {a + b + c} \right)\\
= \left[ {\left( {a + b} \right) + c} \right].\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right).c + {c^2}} \right] - 3ab\left( {a + b + c} \right)\\
= \left( {a + b + c} \right)\left( {{a^2} + 2ab + {b^2} - ac - bc + {c^2}} \right) - 3ab\left( {a + b + c} \right)\\
= \left( {a + b + c} \right)\left( {{a^2} + 2ab + {b^2} - ac - bc + {c^2} - 3ab} \right)\\
= \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)
\end{array}\)
