Question-153046

Question Number 153046 by Tawa11 last updated on 04/Sep/21

Commented by Tawa11 last updated on 04/Sep/21

If real numbers x and y satisfies the system shown, find the value of x^3 + y^3

$$\mathrm{If}\:\mathrm{real}\:\mathrm{numbers}\:\:\mathrm{x}\:\:\mathrm{and}\:\:\mathrm{y}\:\:\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{system}\:\mathrm{shown},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}^{\mathrm{3}} \:\:\:+\:\:\:\mathrm{y}^{\mathrm{3}} \\ $$

Answered by mr W last updated on 04/Sep/21

xy+(5−x)(5−y)+2(√(xy(5−x)(5−y)))=16 25−5x−5y+2xy+2(√(xy(5−x)(5−y)))=16 x(5−y)+y(5−x)+2(√(xy(5−x)(5−y)))=25 5x+5y−2xy+2(√(xy(5−x)(5−y)))=25 10x+10y−4xy−25=9 5(x+y)=2xy+17 (√(xy))+(√(25−5(x+y)+xy))=4 (√(xy))+(√(8−xy))=4 let u=(√(xy)) u+(√(8−u^2 ))=4 8−u^2 =16−8u+u^2 4−4u+u^2 =0 ⇒u=2=(√(xy)) ⇒xy=4 ⇒x+y=((2xy+17)/5)=5 ⇒(x,y)=(1,4) or (4,1) x^3 +y^3 =(x+y)^3 −3xy(x+y)=5^3 −3×4×5=65

$${xy}+\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{16} \\ $$$$\mathrm{25}−\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{2}{xy}+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{16} \\ $$$${x}\left(\mathrm{5}−{y}\right)+{y}\left(\mathrm{5}−{x}\right)+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{25} \\ $$$$\mathrm{5}{x}+\mathrm{5}{y}−\mathrm{2}{xy}+\mathrm{2}\sqrt{{xy}\left(\mathrm{5}−{x}\right)\left(\mathrm{5}−{y}\right)}=\mathrm{25} \\ $$$$\mathrm{10}{x}+\mathrm{10}{y}−\mathrm{4}{xy}−\mathrm{25}=\mathrm{9} \\ $$$$\mathrm{5}\left({x}+{y}\right)=\mathrm{2}{xy}+\mathrm{17} \\ $$$$\sqrt{{xy}}+\sqrt{\mathrm{25}−\mathrm{5}\left({x}+{y}\right)+{xy}}=\mathrm{4} \\ $$$$\sqrt{{xy}}+\sqrt{\mathrm{8}−{xy}}=\mathrm{4} \\ $$$${let}\:{u}=\sqrt{{xy}} \\ $$$${u}+\sqrt{\mathrm{8}−{u}^{\mathrm{2}} }=\mathrm{4} \\ $$$$\mathrm{8}−{u}^{\mathrm{2}} =\mathrm{16}−\mathrm{8}{u}+{u}^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{4}{u}+{u}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{2}=\sqrt{{xy}} \\ $$$$\Rightarrow{xy}=\mathrm{4} \\ $$$$\Rightarrow{x}+{y}=\frac{\mathrm{2}{xy}+\mathrm{17}}{\mathrm{5}}=\mathrm{5} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},\mathrm{4}\right)\:{or}\:\left(\mathrm{4},\mathrm{1}\right) \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{5}^{\mathrm{3}} −\mathrm{3}×\mathrm{4}×\mathrm{5}=\mathrm{65} \\ $$

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

to stay real we need 0≤x≤5∧0≤y≤5 (√(x(5−y)))+(√(y(5−x)))=5 squaring & transforming 2(√(xy(x−5)(y−5)))=2xy−5x−5y+25 squaring & teansforming (x+y−5)^2 =0 ⇒ y=5−x inserting { ((2(√(x(5−x)))=4 ⇒ x=1∨x=4)),((x+5−x=5 (true))) :}

$$\mathrm{to}\:\mathrm{stay}\:\mathrm{real}\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{5}\wedge\mathrm{0}\leqslant{y}\leqslant\mathrm{5} \\ $$$$\sqrt{{x}\left(\mathrm{5}−{y}\right)}+\sqrt{{y}\left(\mathrm{5}−{x}\right)}=\mathrm{5} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming} \\ $$$$\mathrm{2}\sqrt{{xy}\left({x}−\mathrm{5}\right)\left({y}−\mathrm{5}\right)}=\mathrm{2}{xy}−\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{25} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{teansforming} \\ $$$$\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${y}=\mathrm{5}−{x} \\ $$$$\mathrm{inserting} \\ $$$$\begin{cases}{\mathrm{2}\sqrt{{x}\left(\mathrm{5}−{x}\right)}=\mathrm{4}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{4}}\\{{x}+\mathrm{5}−{x}=\mathrm{5}\:\left(\mathrm{true}\right)}\end{cases} \\ $$

Commented by Tawa11 last updated on 04/Sep/21

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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