Question Number 191389 by MATHEMATICSAM last updated on 23/Apr/23
$$\mathrm{Factorize}: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{11}}{\mathrm{8}} \\ $$
Answered by Frix last updated on 23/Apr/23
![x^3 +(1/x^3 )+((11)/8) =^([t=x+(1/x)]) t^3 −3t+((11)/8)= =(t−(1/2))(t+((1−3(√5))/4))(t+((1+3(√5))/4))= =(x+(1/x)−(1/2))(x+(1/x)+((1−3(√5))/4))(x+(1/x)+((1+3(√5))/4))](https://www.tinkutara.com/question/Q191391.png)
$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{11}}{\mathrm{8}}\:\overset{\left[{t}={x}+\frac{\mathrm{1}}{{x}}\right]} {=}\:{t}^{\mathrm{3}} −\mathrm{3}{t}+\frac{\mathrm{11}}{\mathrm{8}}= \\ $$$$=\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}+\frac{\mathrm{1}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)\left({t}+\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)= \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)\left({x}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\right) \\ $$