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

Commented by Tawa11 last updated on 04/Sep/21

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

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

$$xy+(5-x)(5-y)+2\sqrt{xy(5-x)(5-y)}=16$$$$25-5x-5y+2xy+2\sqrt{xy(5-x)(5-y)}=16$$$$x(5-y)+y(5-x)+2\sqrt{xy(5-x)(5-y)}=25$$$$5x+5y-2xy+2\sqrt{xy(5-x)(5-y)}=25$$$$10x+10y-4xy-25=9$$$$5(x+y)=2xy+17$$$$\sqrt{xy}+\sqrt{25-5(x+y)+xy}=4$$$$\sqrt{xy}+\sqrt{8-xy}=4$$$$letu=\sqrt{xy}$$$$u+\sqrt{8-u^2}=4$$$$8-u^2=16-8u+u^2$$$$4-4u+u^2=0$$$$\Rightarrow u=2=\sqrt{xy}$$$$\Rightarrow xy=4$$$$\Rightarrow x+y=\frac{2xy+17}{5}=5$$$$\Rightarrow (x,y)=(1,4)or(4,1)$$$$x^3+y^3={(x+y)}^3-3xy(x+y)=5^3-3\times 4\times 5=65$$

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

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

Commented by Tawa11 last updated on 04/Sep/21

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

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