Đáp án:

$\begin{array}{l}
1){x^3} + 3{x^2} + 3x + 1 = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^3} = 0\\
 \Leftrightarrow x + 1 = 0\\
 \Leftrightarrow x =  - 1\\
Vậy\,x =  - 1\\
2)8{x^3} - 12{x^2} + 6x - 1 = 0\\
 \Leftrightarrow {\left( {2x} \right)^3} - 3.4{x^2}.1 + 3.2x.1 - 1 = 0\\
 \Leftrightarrow {\left( {2x - 1} \right)^3} = 0\\
 \Leftrightarrow 2x - 1 = 0\\
 \Leftrightarrow x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}\\
3)\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) - 8x\left( {{x^2} + 2} \right) = 17\\
 \Leftrightarrow {\left( {2x} \right)^3} + {1^3} - 8{x^3} - 16x = 17\\
 \Leftrightarrow 16x =  - 16\\
 \Leftrightarrow x =  - 1\\
Vậy\,x =  - 1\\
4)\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\\
 \Leftrightarrow 2x =  - 23\\
 \Leftrightarrow x = \dfrac{{ - 23}}{2}\\
Vậy\,x = \dfrac{{ - 23}}{2}\\
5){\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} - x = 28\\
 \Leftrightarrow 3{x^2} + 26x = 0\\
 \Leftrightarrow x\left( {3x + 26} \right) = 0\\
 \Leftrightarrow x = 0;x = \dfrac{{ - 26}}{3}\\
Vậy\,x = 0;x =  - \dfrac{{26}}{3}\\
6){\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)\\
 + 6{\left( {x + 1} \right)^2} = 49\\
 \Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - {x^3} + 27\\
 + 6\left( {{x^2} + 2x + 1} \right) = 49\\
 \Leftrightarrow  - 6{x^2} + 12x + 19 + 6{x^2} + 12x + 6 = 49\\
 \Leftrightarrow 24x = 24\\
 \Leftrightarrow x = 1\\
Vậy\,x = 1\\
7){\left( {{x^2} - 1} \right)^3} - \left( {{x^4} + {x^2} + 1} \right)\left( {{x^2} - 1} \right) = 0\\
 \Leftrightarrow {x^6} - 3{x^4} + 3{x^2} - 1 - {x^6} + 1 = 0\\
 \Leftrightarrow  - 3{x^4} + 3{x^2} = 0\\
 \Leftrightarrow 3{x^2}\left( { - {x^2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
{x^2} = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1\\
x =  - 1
\end{array} \right.\\
Vậy\,x =  - 1;x = 0;x = 1\\
8){\left( {x + 1} \right)^3} + 3{\left( {x + 1} \right)^2}\left( {x + 2} \right)\\
 + 3\left( {x + 1} \right){\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3} = 0\\
 \Leftrightarrow {\left( {x + 1 + x + 2} \right)^3} = 0\\
 \Leftrightarrow 2x + 3 = 0\\
 \Leftrightarrow x =  - \dfrac{3}{2}\\
Vậy\,x = \dfrac{{ - 3}}{2}
\end{array}$