1. Find all real numbers $x$ such that $\left[x^{3}\right]=4 x+3$. Here $[y]$ denotes the greatest integer not exceeding the real number $y$.
（Yang Wenpeng）
1. Let $x$ be a real number that satisfies the condition, then we can set $x=\frac{k}{4}$ $(k \in \mathbf{Z})$, and $\left[\frac{k^{3}}{64}\right]=k+3$. Therefore, $3 \leqslant \frac{k^{3}}{64}-k256$, which is a contradiction.
Hence, $k \in\{-7,-6, \cdots,-1,9\}$.
By calculating each, we find that the $k$ satisfying (1) are only two, i.e., $k=-4,-5$.
Thus, $x=-\frac{5}{4}$ or -1.