Đáp án:

\(\left[ \begin{array}{l}
x = 1\\
x =  - 2\\
x =  - \dfrac{1}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{*{20}{l}}
{{x^3} - 3{x^2} + 3x - 1 + {x^3} + 3.{x^2}.2 + 3.x.4 + 8 = 8{x^3} + 3.4{x^2}.1 + 3.2x.1 + 1}\\
{ \to 2{x^3} + 3{x^2} + 15x + 7 = 8{x^3} + 12{x^2} + 6x + 1}\\
{ \to 6{x^3} + 9{x^2} - 9x - 6 = 0}\\
\begin{array}{l}
 \to 2{x^3} + 3{x^2} - 3x - 2 = 0\\
 \to 2{x^3} - 2{x^2} + 5{x^2} - 5x + 2x - 2 = 0
\end{array}\\
\begin{array}{l}
 \to 2{x^2}\left( {x - 1} \right) + 5x\left( {x - 1} \right) + 2\left( {x - 1} \right) = 0\\
 \to \left( {x - 1} \right)\left( {2{x^2} + 5x + 2} \right) = 0\\
 \to \left( {x - 1} \right)\left( {2{x^2} + 4x + x + 2} \right) = 0\\
 \to \left( {x - 1} \right)\left( {2x\left( {x + 2} \right) + \left( {x + 2} \right)} \right) = 0\\
 \to \left( {x - 1} \right)\left( {x + 2} \right)\left( {2x + 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - 2\\
x =  - \dfrac{1}{2}
\end{array} \right.
\end{array}
\end{array}\)

Đáp án `+` Giải thích các bước giải `!`

`(x-1)^3+(x+2)^3 = (2x+1)^3`

`<=> x^3-3x^2+3x-1+x^3+6x^2+12x+8 = 8x^3+12x^2+6x+1`

`<=> (x^3+x^3)+(-3x^2+6x^2)+(3x+12x)+(-1+8)-8x^3-12x^2-6x-1 = 0`

`<=> 2x^3+3x^2+15x+7-8x^3-12x^2-6x-1 = 0`

`<=> (2x^3-8x^3)+(3x^2-12x^2)+(15x-6x)+(7-1) = 0`

`<=> -6x^3-9x^2+9x+6 = 0`

`<=> -3(2x^3+3x^2-3x-2) = 0`

`<=> 2x^3+3x^2-3x-2 = 0`

`<=> (2x^3-2)+(3x^2-3x) = 0`

`<=> 2(x^3-1)+3x(x-1) = 0`

`<=> 2(x-1)(x^2+x+1)+3x(x-1) = 0`

`<=> (x-1)(2x^2+2x+2+3x) = 0`

`<=> (x-1)(2x^2+5x+2) = 0`

`<=> (x-1)(2x^2+4x+x+2) = 0`

`<=> (x-1)[(2x^2+4x)+(x+2)] = 0`

`<=> (x-1)[2x(x+2)+(x+2)] = 0`

`<=> (x-1)(2x+1)(x+2) = 0`

`<=>` \(\left[ \begin{array}{l}x-1=0\\2x+1=0\\x+2=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1\\2x=-1\\x=-2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{2}\\x=-2\end{array} \right.\) 

Vậy `S= {1; -(1)/2; -2}`