Đáp án:
$\begin{array}{l}
30\\
a){\left( {{x^2} - 1} \right)^3} - \left( {{x^4} + {x^2} + 1} \right)\left( {{x^2} - 1} \right)\\
= {x^6} - 3{x^4} + 3{x^2} - 1 - \left( {{x^6} - 1} \right)\\
= {x^6} - 3{x^4} + 3{x^2} - 1 - {x^6} + 1\\
= 3{x^2} - 3{x^4}\\
b)\left( {{x^4} - 3{x^2} + 9} \right)\left( {{x^2} + 3} \right) - {\left( {3 + {x^2}} \right)^3}\\
= {x^6} + {3^3} - \left( {27 + 27{x^2} + 9{x^4} + {x^6}} \right)\\
= - 27{x^2} - 9{x^4}\\
c){\left( {x - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 6{\left( {x + 1} \right)^2}\\
= {x^3} - 9{x^2} + 27x - 27 - {x^3} + 27\\
+ 6\left( {{x^2} + 2x + 1} \right)\\
= - 9{x^2} + 27x + 6{x^2} + 12x + 6\\
= - 3{x^2} + 39x + 6\\
31)a)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} + 2} \right) = 15\\
\Leftrightarrow {x^3} + 8 - {x^3} - 2x - 15 = 0\\
\Leftrightarrow - 2x - 7 = 0\\
\Leftrightarrow x = - \dfrac{7}{2}\\
Vậy\,x = \dfrac{{ - 7}}{2}\\
b){\left( {x + 3} \right)^3} - x{\left( {3x + 1} \right)^2} + \left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) = 28\\
\Leftrightarrow {x^3} + 9{x^2} + 27x + 27\\
- x\left( {9{x^2} + 6x + 1} \right) + 8{x^3} + 1 = 28\\
\Leftrightarrow 9{x^3} + 9{x^2} + 27x + 28 - 9{x^3} - 6{x^2} - 1 = 28\\
\Leftrightarrow 3{x^2} + 27x = 0\\
\Leftrightarrow 3x\left( {x + 9} \right) = 0\\
\Leftrightarrow x = 0;x = - 9\\
Vậy\,x = 0;x = - 9\\
c){\left( {{x^2} - 1} \right)^3} - \left( {{x^4} + {x^2} + 1} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left[ {{{\left( {{x^2} - 1} \right)}^2} - {x^4} - {x^2} - 1} \right] = 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left( {{x^4} - 2{x^2} + 1 - {x^4} - {x^2} - 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left( { - 3{x^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1\\
x = 0
\end{array} \right.\\
Vậy\,x = 0;x = 1;x = - 1
\end{array}$
