Đáp án:
\(\left[ \begin{array}{l}
x = 0\\
x = 1\\
x = \frac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{{x^3} + {{(x - 1)}^3} = {{(2x - 1)}^3}}\\
{ \to {x^3} + {x^3} - 3{x^2} + 3x - 1 = 8{x^3} - 3.4{x^2} + 3.2x - 1}\\
{ \to 6{x^3} - 9{x^2} + 3x = 0}\\
{ \to 3x\left( {2{x^2} - 3x + 1} \right) = 0}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{2{x^2} - 3x + 1 = 0}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{2{x^2} - 2x - x + 1 = 0}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{2x\left( {x - 1} \right) - \left( {x - 1} \right) = 0}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x - 1 = 0}\\
{2x - 1 = 0}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x = 1}\\
{x = \frac{1}{2}}
\end{array}} \right.}
\end{array}\)
