6. In the geometric sequence $\left\{a_{n}\right\}$, if for any positive integer $n, a_{1}+a_{2}+\cdots+a_{n}=2^{n}-1$, then $a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=$ $\qquad$
6. $\frac{1}{7}\left(8^{n}-1\right)$.

From the conditions, we know $a_{1}=1, a_{2}=2$, common ratio $q=2, a_{n}=2^{n-1}$, hence $a_{n}^{3}=8^{n-1}$ is a geometric sequence.
$$
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=\frac{1}{7}\left(8^{n}-1\right) .
$$