246. Solve the equation $x^{3}-[x]=3$, where, as usual, $[x]$ is the integer part of the number $x$ (see p. 37).
246. Let $\{x\}=x-[x]$ denote the so-called fractional part of the number $x$ (see p. 37); it is clear that $0 \leqslant\{x\}<1$; for $x<0$ and $x^{3}-x=x\left(x^{2}-1\right)<0<2$; for $x=-1$ we have $x^{3}-$ $-x=0<2$; for $-1<x \leqslant 0$ we have $x^{3}-x \leqslant-x<1$; for $0<x \leqslant 1$ we have $x^{3}-x<x^{3} \leqslant 1$. Therefore, it must be $1<x<2$ and, hence, $[x]=1$. Then the original equation takes the form:

$$
x^{3}-1=3, \quad \text { from which } \quad x^{3}=4 \text { and } \quad x=\sqrt[3]{4}
$$

- this is the (unique) solution to the problem.