Question Number 221585 by mr W last updated on 08/Jun/25
$${solve}\:{for}\:{x}\:\in{R} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} ={x}+\mathrm{6} \\ $$
Commented by mr W last updated on 08/Jun/25
$${please}\:{also}\:{solve} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{5}\right)^{\mathrm{3}} ={x}+\mathrm{5} \\ $$
Answered by fantastic last updated on 08/Jun/25
$$\mathrm{2}?? \\ $$
Answered by fantastic last updated on 08/Jun/25
$$\mathrm{2}.{But}\:{i}\:{dont}\:{knpw}\:{how} \\ $$
Answered by Frix last updated on 08/Jun/25
$$\mathrm{Obviously}\:{x}=\mathrm{2} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} \\ $$$${g}\left({x}\right)={x}+\mathrm{6} \\ $$$${f}\:'\left({x}\right)=\mathrm{9}{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{2}} \geqslant\mathrm{0}\forall{x}\in\mathbb{R} \\ $$$$\Rightarrow\:{f}\left({x}\right)\:\mathrm{increasing}\:\forall{x}\in\mathbb{R} \\ $$$$\mathrm{it}\:\mathrm{has}\:\mathrm{flat}\:\mathrm{points}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{216}}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{\sqrt[{\mathrm{3}}]{\mathrm{6}}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{but} \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:{g}\left({x}\right)>{f}\left({x}\right) \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution} \\ $$
Commented by mr W last updated on 08/Jun/25
$${thanks}! \\ $$
Commented by mr W last updated on 08/Jun/25
$${i}\:{should}\:{change}\:{the}\:{question}\:{such} \\ $$$${that}\:{the}\:{solution}\:{is}\:{not}\:{obvious}. \\ $$
Answered by Frix last updated on 08/Jun/25
$$\left({x}^{\mathrm{3}} −{a}\right)^{\mathrm{3}} ={x}+{a} \\ $$$$\mathrm{Let}\:{x}^{\mathrm{3}} ={t}+{a} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{3}} ={x}+{a} \\ $$$$\Rightarrow \\ $$$${t}={x} \\ $$$$\Rightarrow \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\left({x}^{\mathrm{3}} −{a}\right)^{\mathrm{3}} ={x}+{a}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{of}\:{x}^{\mathrm{3}} −{x}−{a}=\mathrm{0} \\ $$
Commented by fantastic last updated on 08/Jun/25
$$\:{wow} \\ $$
Commented by mr W last updated on 08/Jun/25
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Answered by mr W last updated on 08/Jun/25
$${x}^{\mathrm{3}} −\mathrm{5}=\sqrt[{\mathrm{3}}]{{x}+\mathrm{5}} \\ $$$${let}\:{f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{5} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{{f}\left({x}\right)+\mathrm{5}}\:\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{{x}+\mathrm{5}} \\ $$$${the}\:{original}\:{equation}\:{becomes} \\ $$$${f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\: \\ $$$$\Rightarrow{f}\left({x}\right)={x} \\ $$$${x}^{\mathrm{3}} −\mathrm{5}={x}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{x}−\mathrm{5}=\mathrm{0}\: \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{2013}}}{\mathrm{18}}+\frac{\mathrm{5}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{2013}}}{\mathrm{18}}−\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\approx\mathrm{1}.\mathrm{904160859} \\ $$
Answered by mr W last updated on 08/Jun/25
$$\left({x}^{\mathrm{3}} −\mathrm{5}\right)^{\mathrm{3}} ={x}+\mathrm{5} \\ $$$${let}\:{t}={x}^{\mathrm{3}} −\mathrm{5} \\ $$$$\Rightarrow{t}^{\mathrm{3}} ={x}+\mathrm{5}={x}−{t}+{x}^{\mathrm{3}} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{x}^{\mathrm{3}} ={x}−{t} \\ $$$$\Rightarrow\left({t}−{x}\right)\left({t}^{\mathrm{2}} +{xt}+{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +{xt}+{x}^{\mathrm{2}} +\mathrm{1}=\left({t}−\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{t}−{x}=\mathrm{0}\:\Rightarrow{t}={x} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{5}={x} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{x}−\mathrm{5}=\mathrm{0} \\ $$$$……\:{see}\:{above} \\ $$