Question Number 219135 by Rojarani last updated on 20/Apr/25
Answered by Frix last updated on 20/Apr/25
$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \underset{={c}^{\frac{\mathrm{1}}{\mathrm{3}}} } {\underbrace{\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}}={c} \\ $$$$\mathrm{27}{abc}=−\left({a}+{b}−{c}\right)^{\mathrm{3}} \\ $$$$\mathrm{Inserting}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{5832}\left(\mathrm{56}{x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}\left(\mathrm{8}{x}−\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{14}}}{\mathrm{28}} \\ $$