6. If the equation
$$
(x-4)^{2} x^{2}-a(|x-2|+2)=b
$$

has five roots, then the range of values for $a$ is $\qquad$
6. $-\frac{32 \sqrt{6}}{9}<a<0$.

Let $t=x-2$.
Then the equation about $t$
$$
(t-2)^{2}(t+2)^{2}-a(|t|+2)=b
$$

has five roots.
$$
\begin{array}{l}
\text { By } f(t)=(t-2)^{2}(t+2)^{2}-a(|t|+2) \\
=\left(t^{2}-4\right)^{2}-a(|t|+2)
\end{array}
$$

is an even function, so
$$
\begin{array}{l}
b=f(0)=16-2 a . \\
\text { Hence }(t-2)^{2}(t+2)^{2}-a(|t|+2) \\
=16-2 a .
\end{array}
$$

Thus, $t^{4}-8 t^{2}-a|t|=0$ has five distinct real roots, which is equivalent to $t^{4}-8 t^{2}-a|t|=0$ having two positive roots, i.e., $t^{3}-8 t=a$ has two positive roots.
Therefore, $-\frac{32 \sqrt{6}}{9}<a<0$.