Shapovalov A.V.

Solve the equation $\left\{(x+1)^{3}\right\}=x^{3}$.

#
From the definition of the fractional part, it follows that $0 \leq x^{3}<1$, that is, $0 \leq x<1$. The equality $\{a+b\}=a$ holds if and only if $b$ is an integer. Therefore, the number $n=3 x^{2}+3 x$ is an integer. The function $y=3 x^{3}+3 x$ is increasing on $[0,1], y(0)=0, y(1)=6$, so $0 \leq n<6$, and from the roots of the equation $3 x^{2}+3 x-n=0$ the one lying in the interval $[0,1)$ is chosen.

## Answer

$0, \frac{\sqrt{21}-3}{6}, \frac{\sqrt{33}-3}{6}, \frac{\sqrt{5}-1}{2}, \frac{\sqrt{57}-3}{6}, \frac{\sqrt{69}-3}{6}$.