Question Number 221647 by Jubr last updated on 09/Jun/25
$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$
Commented by Frix last updated on 09/Jun/25
$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}−\mathrm{4096}=\mathrm{0} \\ $$$$\mathrm{Substitute}\:{x}={t}−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:\mathrm{use}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{Formula} \\ $$