Question Number 153046 by Tawa11 last updated on 04/Sep/21
Commented by Tawa11 last updated on 04/Sep/21
$$\mathrm{If}\:\mathrm{real}\:\mathrm{numbers}\:\:\mathrm{x}\:\:\mathrm{and}\:\:\mathrm{y}\:\:\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{system}\:\mathrm{shown},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}^{\mathrm{3}} \:\:\:+\:\:\:\mathrm{y}^{\mathrm{3}} \\ $$
Answered by mr W last updated on 04/Sep/21
$${xy}+\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{16} \\ $$$$\mathrm{25}−\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{2}{xy}+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{16} \\ $$$${x}\left(\mathrm{5}−{y}\right)+{y}\left(\mathrm{5}−{x}\right)+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{25} \\ $$$$\mathrm{5}{x}+\mathrm{5}{y}−\mathrm{2}{xy}+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{25} \\ $$$$\mathrm{10}{x}+\mathrm{10}{y}−\mathrm{4}{xy}−\mathrm{25}=\mathrm{9} \\ $$$$\mathrm{5}\left({x}+{y}\right)=\mathrm{2}{xy}+\mathrm{17} \\ $$$$\sqrt{{xy}}+\sqrt{\mathrm{25}−\mathrm{5}\left({x}+{y}\right)+{xy}}=\mathrm{4} \\ $$$$\sqrt{{xy}}+\sqrt{\mathrm{8}−{xy}}=\mathrm{4} \\ $$$${let}\:{u}=\sqrt{{xy}} \\ $$$${u}+\sqrt{\mathrm{8}−{u}^{\mathrm{2}} }=\mathrm{4} \\ $$$$\mathrm{8}−{u}^{\mathrm{2}} =\mathrm{16}−\mathrm{8}{u}+{u}^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{4}{u}+{u}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{2}=\sqrt{{xy}} \\ $$$$\Rightarrow{xy}=\mathrm{4} \\ $$$$\Rightarrow{x}+{y}=\frac{\mathrm{2}{xy}+\mathrm{17}}{\mathrm{5}}=\mathrm{5} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},\mathrm{4}\right)\:{or}\:\left(\mathrm{4},\mathrm{1}\right) \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{5}^{\mathrm{3}} −\mathrm{3}×\mathrm{4}×\mathrm{5}=\mathrm{65} \\ $$
Commented by Tawa11 last updated on 04/Sep/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Answered by MJS_new last updated on 04/Sep/21
$$\mathrm{to}\:\mathrm{stay}\:\mathrm{real}\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{5}\wedge\mathrm{0}\leqslant{y}\leqslant\mathrm{5} \\ $$$$\sqrt{{x}\left(\mathrm{5}−{y}\right)}+\sqrt{{y}\left(\mathrm{5}−{x}\right)}=\mathrm{5} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$$\mathrm{2}\sqrt{{xy}\left({x}−\mathrm{5}\right)\left({y}−\mathrm{5}\right)}=\mathrm{2}{xy}−\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{25} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{teansforming} \\ $$$$\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${y}=\mathrm{5}−{x} \\ $$$$\mathrm{inserting} \\ $$$$\begin{cases}{\mathrm{2}\sqrt{{x}\left(\mathrm{5}−{x}\right)}=\mathrm{4}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{4}}\\{{x}+\mathrm{5}−{x}=\mathrm{5}\:\left(\mathrm{true}\right)}\end{cases} \\ $$
Commented by Tawa11 last updated on 04/Sep/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$