6. Let the function $f(x)=(2 a-b) x+2 b-8$ $(b \leqslant 6)$, when $x \in[-4,2]$, $f(x) \geqslant 0$ always holds. Then the range of $\frac{a^{4}-2 b^{4}}{a^{3} b}$ is $\qquad$ .
6. $\left[-\frac{161}{3},-\frac{39071}{4116}\right]$.

From the given, we have
$$
\left\{\begin{array}{l}
f(-4) \geqslant 0, \\
f(2) \geqslant 0, \\
b \leqslant 6,
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
-4 a+3 b-4 \geqslant 0, \\
a \geqslant 2, \\
b \leqslant 6 .
\end{array}\right.
$$

The region that the point $(a, b)$ satisfies is shown in Figure 4. The value of $\frac{b}{a}$ is the slope of the line connecting the point $(a, b)$ and the origin $O$. It is easy to see that
$$
A(2,6) 、 B\left(\frac{7}{2}, 6\right) \text {. }
$$

Then $k_{O B} \leqslant \frac{b}{a} \leqslant k_{O A}$, which means
$$
\frac{12}{7} \leqslant \frac{b}{a} \leqslant 3 \text {. }
$$

Thus, $\frac{a^{4}-2 b^{4}}{a^{3} b}=\frac{a}{b}-2\left(\frac{b}{a}\right)^{3}$.
Let $k=\frac{b}{a}$, then $\frac{a^{4}-2 b^{4}}{a^{3} b}=\frac{1}{k}-2 k^{3}$, where $k \in\left[\frac{12}{7}, 3\right]$.

It is easy to see that $\frac{1}{k}-2 k^{3}$ is a decreasing function on the interval $\left[\frac{12}{7}, 3\right]$, hence we get the range of values.