8.2. (England, 75). Solve the equation

$$
[\sqrt[3]{1}]+[\sqrt[3]{2}]+\ldots+\left[\sqrt[3]{x^{3}-1}\right]=400
$$

in natural numbers.
8.2. Note that the relation $[\sqrt[3]{m}]=k$, where $m, k \in \mathrm{N}$, is equivalent to the inequality $k^{3} \leqslant m \leqslant(k+1)^{3}-1$. The number of natural numbers $m$ satisfying this condition (for a fixed $k$) is $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$. Therefore, the left side of the equation is $\sum_{k=1}^{x-1} S_{k}$, where

$$
S_{k}=k\left(3 k^{2}+3 k+1\right)
$$

Since $S_{k}>0$ for $k \in \mathrm{N}$, $S_{1}=1 \cdot 7 = 7$, $S_{2}=2 \cdot 19 = 38$, $S_{3}=3 \cdot 37 = 111$, $S_{4}=4 \cdot 61 = 244$, and $S_{1}+S_{2}+S_{3}+S_{4}=400$, the original equation has a unique solution in natural numbers: $x=5$.