\(2\left(x^2+x+1\right)^2-7\left(x-1\right)^3=13\left(x^3-1\right)\)
\(\Leftrightarrow2\left(x^2+x+1\right)^2-7\left(x-1\right)\left(x^2+x+1\right)-13\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(x^2+x+1\right)-7\left(x-1\right)-13\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2x^2+2x+2-7x+7-13x+13\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(2x^2-18x+22\right)=0\)
Ta có: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=> \(2x^2-18x+22=0\)
\(\Leftrightarrow x^2-9x+11=0\)
\(\Leftrightarrow\left(x^2-2\cdot x\cdot\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{37}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{9}{2}\right)^2=\dfrac{37}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{9}{2}=\dfrac{\sqrt{37}}{2}\\x-\dfrac{9}{2}=-\dfrac{\sqrt{37}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{37}}{2}\\x=\dfrac{9-\sqrt{37}}{2}\end{matrix}\right.\)
Vậy pt có 2 nghiệm \(x_1=\dfrac{9+\sqrt{37}}{2};x_2=\dfrac{9-\sqrt{37}}{2}\)
