Question Number 195809 by mr W last updated on 11/Aug/23
$${if}\:{x}^{\mathrm{5}} +{x}+\mathrm{1}=\mathrm{0},\:{find}\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =? \\ $$
Commented by jabarsing last updated on 11/Aug/23
$${hello}\:{dear}\:{W}, \\ $$$$\mathrm{1}\:? \\ $$
Commented by jabarsing last updated on 11/Aug/23
$${if}\:{that}\:{is}\:{my}\:{question};\:{I}\:{wana}\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} \:{or} \\ $$$${x}^{\mathrm{2}} −{x}^{\mathrm{3}} \\ $$
Commented by mr W last updated on 11/Aug/23
$${yes}\:{sir}.\:{i}\:{meant}\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} \:{indeed}. \\ $$
Answered by MM42 last updated on 11/Aug/23
$$ \\ $$$${x}^{\mathrm{5}} +{x}+\mathrm{1}=\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\overset{{if}\:\:{x}\in\mathbb{R}} {\Rightarrow}\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =−\mathrm{1}\:\checkmark \\ $$
Commented by mr W last updated on 11/Aug/23
$${great}! \\ $$
Answered by manxsol last updated on 12/Aug/23
$${x}^{\mathrm{5}} +{x}+\mathrm{1}=\mathrm{0}\:\:\Rightarrow{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =? \\ $$$$−−−−−−−−−−−− \\ $$$${E}={x}^{\mathrm{3}} −{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} ={E}+{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} ={Ex}^{\mathrm{2}} +{x}.{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{5}} ={Ex}^{\mathrm{2}} +{x}\left({E}+{x}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{5}} ={Ex}^{\mathrm{2}} +{xE}+{E}+{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} +{x}+\mathrm{1}={E}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$\mathrm{0}=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({E}+\mathrm{1}\right) \\ $$$${E}=−\mathrm{1} \\ $$
Commented by mr W last updated on 12/Aug/23
��