6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$
6. 54.

Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$

and $q>1$.
Let $t=q^{2}-1$. Then
$$
\begin{array}{l}
2 a_{8}+a_{7}=2 a_{2} q^{6}+a_{1} q^{6} \\
=\left(2 a_{2}+a_{1}\right) q^{6} \\
=\frac{8 q^{6}}{q^{2}-1}=\frac{8(t+1)^{3}}{t} \\
=8\left(t^{2}+3 t+3+\frac{1}{t}\right) .
\end{array}
$$

Let $f(t)=8\left(t^{2}+3 t+3+\frac{1}{t}\right)$.
Then $f^{\prime}(t)=8\left(2 t+3-\frac{1}{t^{2}}\right)$
$$
=8 \cdot \frac{(t+1)^{2}(2 t-1)}{t^{2}} \text {. }
$$

Thus, $f(t)$ is a decreasing function on $\left(0, \frac{1}{2}\right)$ and an increasing function on $\left(\frac{1}{2},+\infty\right)$.
Therefore, $f(t)_{\min }=f\left(\frac{1}{2}\right)=54$.