8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$).
8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$

When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end{array}
$$

Therefore, the solution to the equation can only be within the interval $(-3,3)$, and $[x]$ can only take the values $-3, -2, -1, 0, 1, 2$.
If $[x]=-3$, then $x^{3}=-5$, which does not meet the requirement;
If $[x]=-2$, then $x^{3}=-2 \Rightarrow x=-\sqrt[3]{2}$, which meets the requirement;
If $[x]=-1$, then $x^{3}=1$, which does not meet the requirement;
If $[x]=0$, then $x^{3}=4$, which does not meet the requirement;
If $[x]=1$, then $x^{3}=7 \Rightarrow x=\sqrt[3]{7}$, which meets the requirement;
If $[x]=2$, then $x^{3}=10 \Rightarrow x=\sqrt[3]{10}$, which meets the requirement.
Thus, $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}=-2+7+10=15$.