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