Question Number 161391 by cortano last updated on 17/Dec/21
$$\:\:{f}^{\:\mathrm{3}} \left({x}\right)+{x}^{\mathrm{2}} \:{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\forall{x}\in\mathbb{R}\: \\ $$
Commented by Rasheed.Sindhi last updated on 17/Dec/21
$${f}\left({x}\right)\:{is}\:{not}\:{a}\:{polynomial}. \\ $$
Commented by mr W last updated on 18/Dec/21
$${but}\:{if}\:{the}\:{question}\:{is}\:{just}\:{a}\:{little}\:{bit} \\ $$$${different},\:{for}\:{example} \\ $$$${f}^{\mathrm{3}} \left({x}\right)+{x}^{\mathrm{2}} {f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$$${then}\:{f}\left({x}\right)\:{is}\:{no}\:{polynomial}\:{more}. \\ $$$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{6}} }{\mathrm{27}}}+\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:−\sqrt[{\mathrm{3}}]{\sqrt{\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{6}} }{\mathrm{27}}}−\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$
Commented by mr W last updated on 18/Dec/21
$${yes},\:{it}\:{is}!\: \\ $$$${assume}\:{it}\:{is},\:{then}\:{f}\left({x}\right)={ax}+{b}. \\ $$$${comparing}\:{the}\:{first}\:{and}\:{last}\:{term} \\ $$$${it}'{s}\:{easy}\:{to}\:{see}:\:{a}=\mathrm{1},\:{b}=\mathrm{1}.\: \\ $$$${now}\:{check}: \\ $$$${f}\left({x}\right)={x}+\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\:\:\:{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)={x}^{\mathrm{3}} +{x}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−− \\ $$$$\Sigma:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 18/Dec/21
$${You}'{re}\:{right}\:{sir},{I}\:{mistook}. \\ $$