Question Number 108741 by mr W last updated on 18/Aug/20
![Solve x^3 −[x]=3 (x∈R)](https://www.tinkutara.com/question/Q108741.png)
$${Solve}\:{x}^{\mathrm{3}} −\left[{x}\right]=\mathrm{3} \\ $$$$\left({x}\in{R}\right) \\ $$
Answered by Sarah85 last updated on 18/Aug/20
![first, I think 1<x<2 ⇒ [x]=1 ⇒ x=(4)^(1/3) second, proof: x=z+q; z∈Z∧q∈Q∧0≤q<1 (z+q)^3 −z=3 q=−z+((z+3))^(1/3) 0≤q<1 ⇔ z≤((z+3))^(1/3) <z+1 z∈Z ⇒ z=1 ⇒ 1<x<2...](https://www.tinkutara.com/question/Q108743.png)
$$\mathrm{first},\:\mathrm{I}\:\mathrm{think}\:\mathrm{1}<{x}<\mathrm{2} \\ $$$$\Rightarrow\:\left[{x}\right]=\mathrm{1}\:\Rightarrow\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{second},\:\mathrm{proof}: \\ $$$${x}={z}+{q};\:{z}\in\mathbb{Z}\wedge{q}\in\mathbb{Q}\wedge\mathrm{0}\leqslant{q}<\mathrm{1} \\ $$$$\left({z}+{q}\right)^{\mathrm{3}} −{z}=\mathrm{3} \\ $$$${q}=−{z}+\sqrt[{\mathrm{3}}]{{z}+\mathrm{3}} \\ $$$$\mathrm{0}\leqslant{q}<\mathrm{1}\:\Leftrightarrow\:{z}\leqslant\sqrt[{\mathrm{3}}]{{z}+\mathrm{3}}<{z}+\mathrm{1} \\ $$$${z}\in\mathbb{Z}\:\Rightarrow\:{z}=\mathrm{1}\:\Rightarrow\:\mathrm{1}<{x}<\mathrm{2}… \\ $$
Commented by ajfour last updated on 18/Aug/20
$${excellent}\:{way},\:{thanks}. \\ $$
Commented by Sarah85 last updated on 18/Aug/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Sarah85 last updated on 18/Aug/20
![for n∈Z\{0} the equation x^3 −[x]=n has only one solution ⇔ there′s always only one possible value for z satisfying z≤((z+n))^(1/3) <z+1](https://www.tinkutara.com/question/Q108747.png)
$$\mathrm{for}\:{n}\in\mathbb{Z}\backslash\left\{\mathrm{0}\right\}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{3}} −\left[{x}\right]={n}\:\mathrm{has}\:\mathrm{only} \\ $$$$\mathrm{one}\:\mathrm{solution}\:\Leftrightarrow\:\mathrm{there}'\mathrm{s}\:\mathrm{always}\:\mathrm{only}\:\mathrm{one}\:\mathrm{possible} \\ $$$$\mathrm{value}\:\mathrm{for}\:{z}\:\mathrm{satisfying}\:{z}\leqslant\sqrt[{\mathrm{3}}]{{z}+{n}}<{z}+\mathrm{1} \\ $$
Commented by mr W last updated on 19/Aug/20
$${thanks}! \\ $$
Answered by mr W last updated on 19/Aug/20
![say [x]=n ⇒x=n+f with 0≤f<1 (n+f)^3 −n=3 n+3=(n+f)^3 ≥n^3 (n−1)n(n+1)≤3 ⇒n=0^∗ or ⇒n=1 (1+f)^3 −1=3 ⇒(1+f)^3 =4 ⇒f=(4)^(1/3) −1 ⇒x=1+(4)^(1/3) −1=(4)^(1/3) ^∗ with n=0: f^3 =3 ⇒f=(3)^(1/3) >1 ⇒no solution](https://www.tinkutara.com/question/Q108783.png)
$${say}\:\left[{x}\right]={n} \\ $$$$\Rightarrow{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\left({n}+{f}\right)^{\mathrm{3}} −{n}=\mathrm{3} \\ $$$${n}+\mathrm{3}=\left({n}+{f}\right)^{\mathrm{3}} \geqslant{n}^{\mathrm{3}} \\ $$$$\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\leqslant\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{0}\:^{\ast} \:{or} \\ $$$$\Rightarrow{n}=\mathrm{1} \\ $$$$\left(\mathrm{1}+{f}\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{1}+{f}\right)^{\mathrm{3}} =\mathrm{4} \\ $$$$\Rightarrow{f}=\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$$ \\ $$$$\:^{\ast} \:{with}\:{n}=\mathrm{0}: \\ $$$${f}^{\mathrm{3}} =\mathrm{3}\:\Rightarrow{f}=\sqrt[{\mathrm{3}}]{\mathrm{3}}>\mathrm{1}\:\Rightarrow{no}\:{solution} \\ $$