One, (20 points) For positive numbers $a, b$, define the operation * as follows: when $a \leqslant b$, $a * b=a b$; when $b<a<2 b$, $a * b=b^{2}+a-b$; when $a \geqslant 2 b$, $a * b=a^{2}+b-a$. If $3 * x=8$, find the value of $\left[x^{3}-4 x\right]$ (where $[x]$ represents the greatest integer not exceeding $x$).
If $3 \leqslant x$, then $3 * x=3 x \geqslant 3 \cdot 3>8$. This does not satisfy $3 * x=8$.

If $x<3<2 x$, then $3 * x=x^{2}+3-x=8$, which means $x^{2}-x-5=0$.
Thus, $x=\frac{1+\sqrt{21}}{2}$ (taking the positive root) (since $\sqrt{21}<5$ satisfies $x<3<2 x$).

If $3 \geqslant 2 x$, then $3 * x=3^{2}+x-3=6+x \leqslant 6+\frac{3}{2}$ $<8$, which does not satisfy $3 * x=8$.

Therefore, the $x$ that satisfies $3 * x=8$ is $x=\frac{1+\sqrt{21}}{2}$, which satisfies $x^{2}-x-5=0$.
$$
\begin{array}{l}
\text { So, } x^{3}-4 x=x\left(x^{2}-x-5\right)+x^{2}+x=x^{2}+x \\
\quad=x^{2}-x-5+2 x+5=2 x+5 . \\
{\left[x^{3}-4 x\right]=[2 x+5]=[2 x]+5} \\
=[1+\sqrt{21}]+5=10 \text { is the answer. }
\end{array}
$$