Đặt \(x^2-x=t\)
Ta có: \(4\left(x^2-x+1\right)^3=27\left(x^2-x\right)^2\)
\(\Leftrightarrow4\left(t+1\right)^3=27t^2\)
\(\Leftrightarrow4\left(t^3+3t^2+3t+1\right)=27t^2\)
\(\Leftrightarrow4t^3-15t^2+12t+4=0\)
\(\Leftrightarrow4t^3-8t^2-7t^2+14t-2t+4=0\)
\(\Leftrightarrow4t^2\left(t-2\right)-7t\left(t-2\right)-2\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(4t^2-7t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left[4t\left(t-2\right)+t-2\right]=0\)
\(\Leftrightarrow\left(t-2\right)^2\left(4t+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=2\\t=-\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-x=2\\x^2-x=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-x-2=0\\x^2-x+\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)\left(x+1\right)=0\\\left(x-\frac{1}{2}\right)^2=0\end{cases}}\)
Tập nghiệm: \(S=\left\{2;-1;\frac{1}{2}\right\}\)
