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Question Number 197795 by cortano12 last updated on 29/Sep/23

Answered by mr W last updated on 29/Sep/23

$$\left(1\right)$$$$x+y=x^2+y^2$$$${(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$$$$x=\frac{1}{2}+\frac{1}{\sqrt{2}}\cos \theta$$$$y=\frac{1}{2}+\frac{1}{\sqrt{2}}\sin \theta$$$$t=x+y=1+\sin (\frac{\pi }{4}+\theta )=1+\sin \alpha \in [0,2]$$$$x^3+y^3={(x+y)}^3-\frac{3}{2}(x+y)[{(x+y)}^2-x^2-y^2]$$$$x^3+y^3=\frac{1}{2}{(x+y)}^2[3-(x+y)]$$$$x^3+y^3=\frac{1}{2}{t}^2(3-t)\in [0,2]$$$$\left(2\right)$$$$let{2}^x=a\ne 0,2^y=b\ne 0$$$$a+b=a^2+b^2$$$$8^x+8^y=a^3+b^3\in (0,2]$$

Answered by Frix last updated on 29/Sep/23

$$x^2+y^2-x-y=0$$$$\mathrm{Let}x=u-v\wedge y=u+v\wedge v⩾0$$$$2(u^2+v^2-u)=0$$$$v=\sqrt{u-u^2}$$$$\Rightarrow$$$$x+y=x^2+y^2=2u$$$$x^3+y^3=2u^2(3-2u)$$$$\mathrm{If}x,y\in \mathbb{R}\Rightarrow 0⩽u⩽1\Rightarrow 0⩽x^3+y^3⩽2$$

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