Clearly $x\in [0,1)$. Note that $\{x^3+3x^2+3x+1\} = \{x^3+3x^2+3x\}$, so $x^3+3x^2+3x - [x^3+3x^2+3x]=x^3$, yielding $3x^2+3x = [x^3+3x^2+3x]$. So, $3x^2+3x$ is integral and less than $6$. Set $3x^2+3x=n$ where $n<6$ whose only positive solution is $(-3+\sqrt{12n+9})/6$. Any such $x$ is less than 1, so is a valid solution.