\(\Leftrightarrow x^4+x^3+x^2-1=\left(x+1\right)\left(x^3+x-1\right)\)
\(\Rightarrow x=-1\)
\(\Rightarrow x^3+x-1=0\)
\(\Rightarrow x=\frac{\left(\sqrt{31}+\sqrt{27}\right)^{\frac{1}{3}}}{\sqrt[3]{2}\sqrt{3}}-\frac{\sqrt[3]{2}}{\sqrt{3}\left(\sqrt{31}+\sqrt{27}\right)^{\frac{1}{3}}}\)
vậy x có 2 TH
TH1:x=-1
TH2:x=\(\Rightarrow x=\frac{\left(\sqrt{31}+\sqrt{27}\right)^{\frac{1}{3}}}{\sqrt[3]{2}\sqrt{3}}-\frac{\sqrt[3]{2}}{\sqrt{3}\left(\sqrt{31}+\sqrt{27}\right)^{\frac{1}{3}}}\)