\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1\\ \Leftrightarrow4x^2-4x+x-1=0\\ \Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
