3. In the geometric sequence $\left\{a_{n}\right\}$ with all terms being positive, if $a_{3}+a_{4}-a_{2}-a_{1}=8$, then the minimum value of $a_{5}+a_{6}+a_{7}+a_{8}$ is $\qquad$
$3.44+20 \sqrt{5}$.
Let the common ratio be $q$. From the problem, we know $q>1$.
From $a_{3}+a_{4}-a_{2}-a_{1}=8$, we get
$a_{1}(q+1)\left(q^{2}-1\right)=8$.
And $a_{5}+a_{6}+a_{7}+a_{8}=a_{1}(q+1) q^{4}\left(q^{2}+1\right)$
$$
=\frac{8 q^{4}\left(q^{2}+1\right)}{q^{2}-1} \text {. }
$$

Let $t=q^{2}-1$. Then $t>0$, and
$$
\begin{array}{l}
\frac{8 q^{4}\left(q^{2}+1\right)}{q^{2}-1}=\frac{8(t+1)^{2}(t+2)}{t} \\
=8\left(t^{2}+4 t+5+\frac{2}{t}\right) .
\end{array}
$$

Using derivatives, we can find that the minimum value of the above expression is $44+20 \sqrt{5}$.