Condition : \(\left\{{}\begin{matrix}x^3-1\ne0\\x^2+x+1\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3\ne1\\x^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ne0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ne0\end{matrix}\right.\)\(\Rightarrow x\ne1\)

We have : x³ - 1 = (x - 1)(x² + x + 1)

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

\(\Leftrightarrow\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)+6\left(x^2+x+1\right)}{x^3-1}-6=0\)

\(\Leftrightarrow\dfrac{4x^2-3x+17+2x^2-3x+1+6x^2+6x+6-6x^3+6}{x^3-1}=0\)

\(\Leftrightarrow-6x^3+12x^2+30=0\)\(\Leftrightarrow x^3-2x^2-5=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(\sqrt[3]{\dfrac{151+3\sqrt{2505}}{16}}+\sqrt[3]{\dfrac{151-3\sqrt{2505}}{16}}+1\right)\)

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