\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow MTC\) là \(\left(x-1\right)\left(x^2+x+1\right)\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)