Question Number 84515 by M±th+et£s last updated on 13/Mar/20
$${Q}.{solve} \\ $$$${x}^{\mathrm{3}} −{x}={x}! \\ $$
Answered by mr W last updated on 13/Mar/20
$$\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)! \\ $$$${x}\neq\mathrm{0},\:{x}\neq\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{1}=\left({x}−\mathrm{2}\right)! \\ $$$${let}\:{t}={x}−\mathrm{2} \\ $$$$\Rightarrow{t}+\mathrm{3}={t}! \\ $$$$\Rightarrow{t}=\mathrm{3} \\ $$$$\Rightarrow{x}=\mathrm{3}+\mathrm{2}=\mathrm{5} \\ $$
Commented by mr W last updated on 14/Mar/20
$${at}\:{t}=\mathrm{2}: \\ $$$${t}+\mathrm{3}=\mathrm{5},\:{t}!=\mathrm{2}\:\Rightarrow{t}+\mathrm{3}>{t}! \\ $$$${at}\:{t}=\mathrm{4}: \\ $$$${t}+\mathrm{3}=\mathrm{7},\:{t}!=\mathrm{24}\:\Rightarrow{t}+\mathrm{3}<{t}! \\ $$$${that}\:{means}\:{between}\:{t}=\mathrm{2}\:{and}\:{t}=\mathrm{4}\:{there} \\ $$$${is}\:{a}\:{point}\:{for}\:{t}+\mathrm{3}={t}!.\:{we}\:{can}\:{check} \\ $$$${with}\:{t}=\mathrm{3}:\:{t}+\mathrm{3}=\mathrm{6}={t}!=\mathrm{6}.\:{i}.{e}.\:{t}=\mathrm{3}\:{is} \\ $$$${a}\:{solution}\:{and}\:{the}\:{only}\:{solution}, \\ $$$${because}\:{upon}\:{t}=\mathrm{4},\:{t}!\:\gg{t}+\mathrm{3}. \\ $$
Commented by M±th+et£s last updated on 14/Mar/20
$${thank}\:{you}\:{sir}\: \\ $$$${but}\:{how}\:{did}\:{you}\:{get}\:{that} \\ $$$${t}+\mathrm{3}={t}! \\ $$$${t}=\mathrm{3}\:\:\:\:?? \\ $$
Commented by M±th+et£s last updated on 14/Mar/20
$${god}\:{bless}\:{you}\:{sir} \\ $$