(6) In the geometric sequence $\left\{a_{n}\right\}$ with all terms being positive, $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $\qquad$.
654 Prompt: Let the common ratio of the sequence $\left\{a_{n}\right\}$ be $q(q>0), 2 a_{4}+a_{3}-2 a_{2}-a_{1}=$ 8 can be transformed into
$$
\begin{array}{c}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8, \\
\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8, \\
2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$

and $q>1$, let $t=q^{2}-1$, then $q^{2}=t+1, t>0$.
$$
\begin{aligned}
2 a_{8}+a_{7} & =2 a_{2} q^{6}+a_{1} q^{6}=\left(2 a_{2}+a_{1}\right) q^{6} \\
& =\frac{8 q^{6}}{q^{2}-1}=\frac{8(t+1)^{3}}{t} \\
& =8\left(t^{2}+3 t+3+\frac{1}{t}\right) .
\end{aligned}
$$

Let $f(t)=8\left(t^{2}+3 t+3+\frac{1}{t}\right)$, then
$$
\begin{aligned}
f^{\prime}(t) & =8\left(2 t+3-\frac{1}{t^{2}}\right) \\
& =8\left(\frac{2 t^{3}+3 t^{2}-1}{t^{2}}\right) \\
& =8\left(\frac{(t+1)^{2}(2 t-1)}{t^{2}}\right) .
\end{aligned}
$$

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