Question Number 76307 by Master last updated on 26/Dec/19
Answered by mr W last updated on 26/Dec/19
$${x}\neq\mathrm{1} \\ $$$$\mathrm{125}\left(\mathrm{1}+{x}^{\mathrm{5}} \right)=\mathrm{11}\left(\mathrm{1}+{x}\right)^{\mathrm{5}} \\ $$$$\mathrm{125}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right)=\mathrm{11}\left(\mathrm{1}+{x}\right)^{\mathrm{4}} \\ $$$$\mathrm{114}{x}^{\mathrm{4}} −\mathrm{169}{x}^{\mathrm{3}} +\mathrm{59}{x}^{\mathrm{2}} −\mathrm{169}{x}+\mathrm{114}=\mathrm{0} \\ $$$$\left(\mathrm{3}{x}−\mathrm{2}\right)\left(\mathrm{2}{x}−\mathrm{3}\right)\left(\mathrm{19}{x}^{\mathrm{2}} +\mathrm{13}{x}+\mathrm{19}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{x}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{3},\mathrm{4}} =\frac{−\mathrm{13}\pm{i}\mathrm{5}\sqrt{\mathrm{51}}}{\mathrm{38}} \\ $$
Commented by Master last updated on 26/Dec/19
$$\mathrm{sir}\:\:\mathrm{125}\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} \right)=\mathrm{11}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{5}} \\ $$$$\mathrm{125}\frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} \right)}{\left(\mathrm{1}+\mathrm{x}\right)}=\mathrm{11}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} \:\Leftrightarrow\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} \right)\:??? \\ $$
Commented by john santuy last updated on 26/Dec/19
$$\left(\mathrm{1}+{x}\right)^{\mathrm{5}} =\mathrm{1}+\mathrm{5}{x}+\mathrm{10}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \\ $$$$\mathrm{1}+{x}^{\mathrm{5}} \:=\left(\mathrm{1}+{x}\right)^{\mathrm{5}} −\mathrm{5}{x}−\mathrm{10}{x}^{\mathrm{2}} −\mathrm{10}{x}^{\mathrm{3}} −{x}^{\mathrm{4}} \\ $$
Commented by mr W last updated on 26/Dec/19
$${is}\:{it}\:{a}\:{question}\:{sir}?\:{can}\:{you}\:{please} \\ $$$${use}\:{more}\:{words}\:{to}\:{make}\:{clear} \\ $$$${what}\:{you}\:{mean}\:{instead}\:{of}\:\:{letting} \\ $$$${guess}?\:{thank}\:{you}! \\ $$
Commented by Master last updated on 26/Dec/19
$$\mathrm{thanks} \\ $$
Commented by mr W last updated on 26/Dec/19
$$\mathrm{1}+{x}^{\mathrm{5}} =\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right) \\ $$