Question Number 193687 by Mingma last updated on 18/Jun/23
Commented by Frix last updated on 18/Jun/23
$$\mathrm{Squaring}\:\&\:\mathrm{transforming}\:\mathrm{3}\:\mathrm{times}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{8}} −\frac{\mathrm{10}{x}^{\mathrm{7}} }{\mathrm{3}}+\frac{\mathrm{143}{x}^{\mathrm{6}} }{\mathrm{27}}−\frac{\mathrm{854}{x}^{\mathrm{5}} }{\mathrm{81}}+\frac{\mathrm{4352}{x}^{\mathrm{4}} }{\mathrm{243}}−\frac{\mathrm{4678}{x}^{\mathrm{3}} }{\mathrm{243}}+\frac{\mathrm{611}{x}^{\mathrm{2}} }{\mathrm{27}}−\frac{\mathrm{1922}{x}}{\mathrm{81}}+\frac{\mathrm{817}}{\mathrm{81}}=\mathrm{0} \\ $$$${x}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{factorization}: \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{3}\right)\left({x}^{\mathrm{5}} −\frac{\mathrm{10}{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{\mathrm{143}{x}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{611}{x}^{\mathrm{2}} }{\mathrm{81}}+\frac{\mathrm{1922}{x}}{\mathrm{243}}−\frac{\mathrm{817}}{\mathrm{243}}\right)=\mathrm{0} \\ $$
Commented by Mingma last updated on 18/Jun/23
Perfect ��