Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
a,\\
{x^3} + 6{x^2} + 12x + 8 = {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = {\left( {x + 2} \right)^3}\\
b,\\
{x^3} - 3{x^2} + 3x - 1 = {\left( {x - 1} \right)^3}\\
c,\\
1 - 9x + 27{x^2} - 27{x^3} = 1 - {3.1^2}.3x + 3.1.{\left( {3x} \right)^2} - {\left( {3x} \right)^3} = {\left( {1 - 3x} \right)^3}\\
d,\\
{x^3} + \frac{3}{2}{x^2} + \frac{3}{4}x + \frac{1}{8} = {x^3} + 3.{x^2}.\frac{1}{2} + 3.x.{\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^3} = {\left( {x + \frac{1}{2}} \right)^3}\\
e,\\
27{x^3} - 54{x^2}y + 36x{y^2} - 8{y^3} = {\left( {3x} \right)^3} - 3.{\left( {3x} \right)^2}.2y + 3.3x.{\left( {2y} \right)^2} - {\left( {2y} \right)^3} = {\left( {3x - 2y} \right)^3}\\
5,\\
a,\\
{x^2} - 4{x^2}{y^2} + {y^2} + 2xy = \left( {{x^2} + 2xy + {y^2}} \right) - 4{x^2}{y^2} = {\left( {x + y} \right)^2} - {\left( {2xy} \right)^2} = \left( {x + y - 2xy} \right)\left( {x + y + 2xy} \right)\\
b,\\
{x^6} - {y^6} = {\left( {{x^3}} \right)^2} - {\left( {{y^3}} \right)^2} = \left( {{x^3} - {y^3}} \right)\left( {{x^3} + {y^3}} \right) = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\
c,\\
25 - {a^2} + 2ab - {b^2} = 25 - \left( {{a^2} - 2ab + {b^2}} \right) = {5^2} - {\left( {a - b} \right)^2} = \left( {5 - a + b} \right)\left( {5 + a - b} \right)\\
d,\\
4{b^2}{c^2} - {\left( {{b^2} + {c^2} - {a^2}} \right)^2} = {\left( {2bc} \right)^2} - {\left( {{b^2} + {c^2} - {a^2}} \right)^2}\\
= \left[ {2bc - \left( {{b^2} + {c^2} - {a^2}} \right)} \right].\left[ {2bc + \left( {{b^2} + {c^2} - {a^2}} \right)} \right]\\
= \left[ {{a^2} - \left( {{b^2} - 2bc + {c^2}} \right)} \right].\left[ {\left( {{b^2} + 2bc + {c^2}} \right) - {a^2}} \right]\\
= \left[ {{a^2} - {{\left( {b - c} \right)}^2}} \right].\left[ {{{\left( {b + c} \right)}^2} - {a^2}} \right]\\
= \left( {a - b + c} \right)\left( {a + b - c} \right)\left( {b + c - a} \right)\left( {b + c + a} \right)\\
e,\\
{\left( {a + b + c} \right)^2} + {\left( {a + b - c} \right)^2} - 4{c^2}\\
= \left[ {{{\left( {a + b + c} \right)}^2} - {{\left( {2c} \right)}^2}} \right] + {\left( {a + b - c} \right)^2}\\
= \left( {a + b + c - 2c} \right).\left( {a + b + 2c} \right) + {\left( {a + b - c} \right)^2}\\
= \left( {a + b - c} \right)\left( {a + b + 2c} \right) + {\left( {a + b - c} \right)^2}\\
= \left( {a + b - c} \right).\left[ {\left( {a + b + 2c} \right) + \left( {a + b - c} \right)} \right]\\
= \left( {a + b - c} \right).\left( {2a + 2b + c} \right)\\
1b,\\
3{x^2} + 9x - 30 = 3.\left( {{x^2} + 3x - 10} \right)\\
= 3.\left[ {\left( {{x^2} - 2x} \right) + \left( {5x - 10} \right)} \right]\\
= 3.\left[ {x\left( {x - 2} \right) + 5.\left( {x - 2} \right)} \right]\\
= 3.\left( {x - 2} \right)\left( {x + 5} \right)
\end{array}\)
