Đáp án:
\(\begin{array}{l}
a,\\
{\left( {x + 2} \right)^3}\\
b,\\
{\left( {x - 1} \right)^3}\\
c,\\
{\left( {1 - 3x} \right)^3}\\
d,\\
- \left( {b - c - a} \right).\left( {b - c + a} \right).\left( {b + c - a} \right).\left( {b + c + a} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^3} + 6{x^2} + 12x + 8\\
= {x^3} + 3.2.{x^2} + 3.4.x + 8\\
= {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3}\\
= {\left( {x + 2} \right)^3}\\
b,\\
{x^3} - 3{x^2} + 3x - 1\\
= {x^3} - 3.{x^2}.1 + 3.x{.1^2} - {1^3}\\
= {\left( {x - 1} \right)^3}\\
c,\\
1 - 9x + 27{x^2} - 27{x^3}\\
= 1 - 3.3x + 3.9{x^2} - {3^3}.{x^3}\\
= {1^3} - {3.1^2}.3x + 3.1.{\left( {3x} \right)^2} - {\left( {3x} \right)^3}\\
= {\left( {1 - 3x} \right)^3}\\
d,\\
4{b^2}{c^2} - {\left( {{b^2} + {c^2} - {a^2}} \right)^2}\\
= {\left( {2bc} \right)^2} - {\left( {{b^2} + {c^2} - {a^2}} \right)^2}\\
= \left[ {2bc - \left( {{b^2} + {c^2} - {a^2}} \right)} \right].\left[ {2bc + \left( {{b^2} + {c^2} - {a^2}} \right)} \right]\\
= \left[ {2bc - {b^2} - {c^2} + {a^2}} \right].\left[ {2bc + {b^2} + {c^2} - {a^2}} \right]\\
= - \left[ { - 2bc + {b^2} + {c^2} - {a^2}} \right].\left[ {2bc + {b^2} + {c^2} - {a^2}} \right]\\
= - \left[ {\left( {{b^2} - 2bc + {c^2}} \right) - {a^2}} \right].\left[ {\left( {{b^2} + 2bc + {c^2}} \right) - {a^2}} \right]\\
= - \left[ {{{\left( {b - c} \right)}^2} - {a^2}} \right].\left[ {{{\left( {b + c} \right)}^2} - {a^2}} \right]\\
= - \left[ {\left( {b - c} \right) - a} \right].\left[ {\left( {b - c} \right) + a} \right].\left[ {\left( {b + c} \right) - a} \right].\left[ {\left( {b + c} \right) + a} \right]\\
= - \left( {b - c - a} \right).\left( {b - c + a} \right).\left( {b + c - a} \right).\left( {b + c + a} \right)
\end{array}\)
