Question Number 202964 by ajfour last updated on 06/Jan/24
$${just}\:{the}\:{exact}\:{real}\:{root}\:{please}: \\ $$$$\left({x}^{\mathrm{6}} −\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{2}{x}\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{1}\right)=\mathrm{0} \\ $$