6. Let $0 \leqslant x \leqslant 8$. Then
$$
f(x)=\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1}
$$

the range of values is
6. $[0,4]$.

Notice that
$$
\begin{array}{l}
\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1} \\
=\frac{\sqrt{\left(x^{2}+8\right)\left(8 x-x^{2}\right)}}{x+1} \\
\leqslant \frac{\left(x^{2}+8\right)+\left(8 x-x^{2}\right)}{2(x+1)}=4 .
\end{array}
$$

When $x=2$, the equality holds.