Question Number 192721 by York12 last updated on 25/May/23
$$ \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{18} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} =\mathrm{26} \\ $$$${and}\:{what}\:{do}\:{you}\:{recommend}\:{to}\:{read}\:{to}\:{deal} \\ $$$${with}\:{such}\:{problems} \\ $$
Answered by Frix last updated on 25/May/23
$${y}={px} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} =\frac{\mathrm{18}}{\mathrm{1}−\mathrm{3}{p}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{3}} =\frac{\mathrm{26}}{\mathrm{3}{p}−{p}^{\mathrm{3}} } \\ $$$$\frac{\mathrm{18}}{\mathrm{1}−\mathrm{3}{p}^{\mathrm{2}} }=\frac{\mathrm{26}}{\mathrm{3}{p}−{p}^{\mathrm{3}} } \\ $$$${p}^{\mathrm{3}} −\frac{\mathrm{13}{p}^{\mathrm{2}} }{\mathrm{3}}−\mathrm{3}{p}+\frac{\mathrm{13}}{\mathrm{9}}=\mathrm{0} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{3}}\vee{p}=\mathrm{2}\pm\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} =\mathrm{27}\vee{x}^{\mathrm{3}} =−\frac{\mathrm{27}}{\mathrm{4}}\pm\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\mathrm{The}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy}. \\ $$
Commented by Frix last updated on 25/May/23
$$\mathrm{Sorry}\:\mathrm{I}\:\mathrm{learned}\:\mathrm{this}\:\mathrm{in}\:\mathrm{school}\:\mathrm{years}\:\mathrm{ago}.\:\mathrm{I} \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{recommend}\:\mathrm{any}\:\mathrm{books}. \\ $$
Commented by York12 last updated on 25/May/23
$${okay}\:{sir}\:{thanks}\: \\ $$