Đáp án:

\(\begin{array}{l}
a,\\
 - 3{x^4} + 3{x^2}\\
b,\\
 - 9{x^4} - 27{x^2}\\
c,\\
 - 3{x^2} + 39x + 6
\end{array}\)

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[ {{{\left( {{x^2}} \right)}^2} - 2.{x^2}.1 + {1^2}} \right] - \left( {{x^4} + {x^2} + 1} \right)} \right\}\\
 = \left( {{x^2} - 1} \right).\left[ {\left( {{x^4} - 2{x^2} + 1} \right) - \left( {{x^4} + {x^2} + 1} \right)} \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[ {\left( {{x^4} - 3{x^2} + 9} \right) - \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 - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 6.{\left( {x + 1} \right)^2}\\
 = \left( {x - 3} \right).\left[ {{{\left( {x - 3} \right)}^2} - \left( {{x^2} + 3x + 9} \right)} \right] + 6.\left( {{x^2} + 2.x.1 + {1^2}} \right)\\
 = \left( {x - 3} \right).\left[ {\left( {{x^2} - 2.x.3 + {3^2}} \right) - \left( {{x^2} + 3x + 9} \right)} \right] + 6.\left( {{x^2} + 2x + 1} \right)\\
 = \left( {x - 3} \right).\left( {{x^2} - 6x + 9 - {x^2} - 3x - 9} \right) + 6\left( {{x^2} + 2x + 1} \right)\\
 = \left( {x - 3} \right).\left( { - 9x} \right) + 6{x^2} + 12x + 6\\
 =  - 9{x^2} + 27x + 6{x^2} + 12x + 6\\
 =  - 3{x^2} + 39x + 6
\end{array}\)