Giải phương trình x3+(x-1)3=(2x-1)3

$b)x^3+{\left(x-1\right)}^3={\left(2x-1\right)}^3\iff x^3+{\left(x-1\right)}^3={\left(x+x-1\right)}^3$

Đặt $t=x-\mathrm{1,}$  phương trình thành:

$\begin{array}{l}{x}^3+t^3={\left(x+t\right)}^3\iff x^3+t^3=x^3+t^3+3x^{2}t+3x{t}^2\\ \iff 3xt\left(x+t\right)=0\iff \left[\begin{array}{l}x=0\\ t=0\\ x=-t\end{array}\right.\iff \left[\begin{array}{l}x=0\\ x=1\\ x=\frac{1}{2}\end{array}\right.\end{array}$

Vậy $S=\left\{0;1;\frac{1}{2}\right\}$