Solutions to $\{x^3\}+\lfloor x^4\rfloor=1$ 
Find all solutions of $$\{x^3\}+[x^4]=1$$ 
  where $[x]=\lfloor x\rfloor$

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I know that $0\le\{x^3\}<1\Rightarrow 0<[x^4]\le 1$. Thus $[x^4]=1$. I couldn't get any further though since I'm having trouble with $x^4$ in the term $[x^4]$. $$$$As an example, in another question, I was given the expression $$[2x]-[x+1]$$ In this, to 'convert' the terms inside the floor function, I split the values of $x$ into 2 cases: $x=[x]+\{x\}, \{x\}\in[0,0.5)$ and $x=[x]+\{x\}, \{x\}\in[0.5,1)$. $$$$In this way, I was able to split the value of $[2x]$ into $2[x]$ and $2[x]+1$ respectively. Hence the expressions within the floor function became easier to deal with as the expression $[2x]-[x+1]$ was converted into $2[x]-[x]-1$ and $2[x]+1-[x]-1$ respectively.$$$$
Is there any way to do something similar with $x^4$ in $[x^4]$ so that the $x^3$ inside $\{x^3\}$ and $x^4$ inside $[x^4]$ are converted into the 'same kind'? If this isn't possible, is there any other way to solve this problem, and other problems in which the expressions within the floor/fractional part function, or the floor/fractional part functions themselves are raised to powers?
$$$$Many thanks in anticipation.
 A: Suppose that $\{x^3\}+\lfloor x^4\rfloor=1$. Clearly, $\lfloor x^4\rfloor$ is an integer, so that $$\{x^3\}=1-\lfloor x^4\rfloor$$ must be an integer, too. But the fractional part of any number is always contained in $[0,1)$. This implies that $\{x^3\}=0$, so that $x^3$ is an integer.
Going back to the original equation, one has that $\lfloor x^4\rfloor=1$, so that $x^4\in[1,2)$, implying that $x\in[1,2^{1/4})\cup(-2^{1/4},-1]$. Therefore, $x^3\in[1,2^{3/4})\cup(-2^{3/4},-1]$. But, as observed earlier, $x^3$ is an integer, and since $1<2^{3/4}<2$, the only integers in the set $[1,2^{3/4})\cup(-2^{3/4},-1]$ are $1$ and $-1$. It follows that $x^3\in\{-1,1\}$, or $x\in\{-1,1\}$. These are the only two solutions.
A: I think taking intervals for {$x$} will complicate the question. Maybe it would be easier to think whether or not {$x^3$} is zero or not. If it is not zero, the equation will have no solution.
A: $\{x^3\}=0$ and $\lfloor x^4 \rfloor=1$ implies $$x \in \{\sqrt[3]{n} | n \in \mathbb{Z} \} \cap \left([1, \sqrt[4]{2}) \cup (-\sqrt[4]{2}, -1]\right)=\{\pm 1\}$$, so $x=\pm 1$
