6. Given 2015 positive integers $a_{1}, a_{2}, \cdots, a_{2015}$ satisfying
$$
\begin{array}{l}
a_{1}=1, a_{2}=8, \\
a_{n+1}=3 a_{n}-2 a_{n-1}(n \geqslant 2, \text { and } n \in \mathbf{N}) .
\end{array}
$$

Then the sum of all positive divisors of $a_{2015}-a_{2014}$ is $\qquad$
6. $8\left(2^{2014}-1\right)$.
$$
\begin{array}{l}
\text { Given } a_{n+1}=3 a_{n}-2 a_{n-1} \\
\Rightarrow a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right) . \\
\text { Then } a_{2015}-a_{2014}=2\left(a_{2014}-a_{2013}\right) \\
=2^{2}\left(a_{2013}-a_{2012}\right)=\cdots \\
=2^{2013}\left(a_{2}-a_{1}\right)=7 \times 2^{2013} .
\end{array}
$$

Therefore, the sum of all positive divisors of $a_{2015}-a_{2014}$ is
$$
\frac{7^{2}-1}{7-1} \times \frac{2^{2014}-1}{2-1}=8\left(2^{2014}-1\right) .
$$