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Question Number 108741 by mr W last updated on 18/Aug/20

$$Solve{x}^3-\left[x\right]=3$$$$(x\in R)$$

Answered by Sarah85 last updated on 18/Aug/20

$$\mathrm{first},\mathrm{I}\mathrm{think}1<x<2$$$$\Rightarrow \left[x\right]=1\Rightarrow x=\sqrt[3]{4}$$$$$$$$\mathrm{second},\mathrm{proof}:$$$$x=z+q;z\in \mathbb{Z}\wedge q\in \mathbb{Q}\wedge 0⩽q<1$$$${(z+q)}^3-z=3$$$$q=-z+\sqrt[3]{z+3}$$$$0⩽q<1\iff z⩽\sqrt[3]{z+3}<z+1$$$$z\in \mathbb{Z}\Rightarrow z=1\Rightarrow 1<x<2...$$

Commented by ajfour last updated on 18/Aug/20

$$excellentway,thanks.$$

Commented by Sarah85 last updated on 18/Aug/20

$$\mathrm{thank}\mathrm{you}$$

Commented by Sarah85 last updated on 18/Aug/20

$$\mathrm{for}n\in \mathbb{Z}\mathrm{\setminus }\left\{0\right\}\mathrm{the}\mathrm{equation}{x}^3-\left[x\right]=n\mathrm{has}\mathrm{only}$$$$\mathrm{one}\mathrm{solution}\iff {\mathrm{there}}^{\prime }\mathrm{s}\mathrm{always}\mathrm{only}\mathrm{one}\mathrm{possible}$$$$\mathrm{value}\mathrm{for}z\mathrm{satisfying}z⩽\sqrt[3]{z+n}<z+1$$

Commented by mr W last updated on 19/Aug/20

$$thanks!$$

Answered by mr W last updated on 19/Aug/20

$$say\left[x\right]=n$$$$\Rightarrow x=n+fwith0⩽f<1$$$${(n+f)}^3-n=3$$$$n+3={(n+f)}^3⩾n^3$$$$(n-1)n(n+1)⩽3$$$$\Rightarrow n=0{}^{\ast }or$$$$\Rightarrow n=1$$$${(1+f)}^3-1=3$$$$\Rightarrow {(1+f)}^3=4$$$$\Rightarrow f=\sqrt[3]{4}-1$$$$\Rightarrow x=1+\sqrt[3]{4}-1=\sqrt[3]{4}$$$$$$$${}^{\ast }withn=0:$$$$f^3=3\Rightarrow f=\sqrt[3]{3}>1\Rightarrow nosolution$$

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