Question Number 172632 by Mikenice last updated on 29/Jun/22 | ||
Commented by peter frank last updated on 29/Jun/22 | ||
$$\mathrm{thank}\:\mathrm{you} \\ $$ | ||
Commented by mr W last updated on 29/Jun/22 | ||
$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}} \\ $$$$={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right) \\ $$$$={x}\left(\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {x}+\mathrm{8}\right) \\ $$$$={x}\left(\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right){x}+\mathrm{8}\right) \\ $$$$={x}\left(\left(\mathrm{2}−\mathrm{3}{x}\right){x}+\mathrm{8}\right) \\ $$$$={x}\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}\right) \\ $$$$={x}\left(\mathrm{2}{x}−\mathrm{3}\left(\mathrm{1}−{x}\right)+\mathrm{8}\right) \\ $$$$=\mathrm{5}{x}\left({x}+\mathrm{1}\right) \\ $$$$=\mathrm{5} \\ $$ | ||
Commented by Rasheed.Sindhi last updated on 29/Jun/22 | ||
$$\mathcal{N}{ice}\:\boldsymbol{{sir}}! \\ $$ | ||
Commented by JDamian last updated on 29/Jun/22 | ||
$$={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right)\:\:\:\:\:{shouldn}'{t}\:{be}\:{a}\:\mathrm{3}? \\ $$ | ||
Commented by mr W last updated on 29/Jun/22 | ||
$${why}? \\ $$$${i}\:{did}: \\ $$$${x}^{\mathrm{2}} +{x}=\mathrm{1}\:\Rightarrow{x}\left({x}+\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{1}}={x} \\ $$$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}}={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right) \\ $$ | ||
Commented by Tawa11 last updated on 30/Jun/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||