\(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
= \(\dfrac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
= \(\dfrac{4x^2-3x+17+2x^2-2x-x+1-6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
= \(\dfrac{24}{x^3-1}\)
\(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
=\(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{\left(2x-1\right)\left(x-1\right)}{x^3-1}-\dfrac{6\left(x^2+x+1\right)}{x^3-1}\)
\(=\dfrac{4x^2-3x+17+2x^2-2x-x+1-6x^2-6x-6}{x^3-1}\)\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
