II. Let real numbers $x, y$ satisfy the equation $x^{3}+y^{8}=a^{8}$ $(a>0)$. Find the range of $x+y$.
II. Solution Let $x+y=t$, then $t^{3}=a^{3}+3xyt$ hence $xy=\frac{t^{3}-a^{3}}{3t}$. By Vieta's theorem, $x, y$ are the roots of the equation $z^{2}-tz+\frac{t^{3}-a^{3}}{3t}=0$. Since $x, y$ are real numbers, $\Delta \geqslant 0$, i.e., $t^{2}-4 \cdot \frac{t^{3}-a^{3}}{3t} \geqslant 0$, solving this yields $0<t \leqslant \sqrt[3]{4}a$. Therefore, the range of $x+y$ is $(0, \sqrt[3]{4}a]$.

III. Solution Let $\{x\}=x-[x]$, then $[x]=x-\{x\}$. Substituting into the original equation gives $x^{2}-2x-5=-2\{x\}$.
Since $0 \leqslant \{x\}<1$,
it follows that $-2<x^{2}-2x-5 \leqslant 0$.