6. Let the sequence of positive numbers $x_{1}, x_{2}, \cdots, x_{n}, \cdots$ satisfy
$$
\left(8 x_{2}-7 x_{1}\right) x_{1}^{7}=8
$$

and $x_{k+1} x_{k-1}-x_{k}^{2}=\frac{x_{k-1}^{8}-x_{k}^{8}}{\left(x_{k} x_{k-1}\right)^{7}}(k \geqslant 2)$.
Find the positive real number $a$, such that when $x_{1}>a$, the sequence is monotonically decreasing, i.e., $x_{1}>x_{2}>\cdots>x_{n}>\cdots$; and when $0<x_{1}<a$, the sequence does not have monotonicity.
（Provided by Li Shenghong）
6. From $x_{k+1} x_{k-1}-x_{k}^{2}=\frac{x_{k-1}^{8}-x_{k}^{8}}{\left(x_{k} x_{k-1}\right)^{7}}$, we have
$\frac{x_{k+1}}{x_{k}}-\frac{x_{k}}{x_{k-1}}=\frac{1}{x_{k}^{8}}-\frac{1}{x_{k-1}^{8}}$,
which means $\frac{x_{k+1}}{x_{k}}-\frac{1}{x_{k}^{8}}=\frac{x_{k}}{x_{k-1}}-\frac{1}{x_{k-1}^{8}}=\cdots=\frac{x_{2}}{x_{1}}-\frac{1}{x_{1}^{8}}=\frac{7}{8}$.
Thus, $x_{k+1}=\frac{7}{8} x_{k}+x_{k}^{-7}$.
When $x_{1}>0$, $x_{k}>0(k \geqslant 2)$.
Since $x_{k+1}-x_{k}=x_{k}\left(x_{k}^{-8}-\frac{1}{8}\right)$, when $x_{k}^{-8}-\frac{1}{8}8^{\frac{1}{8}}$, we have $x_{k+1}-x_{k}8^{\frac{1}{8}}$, we have $x_{1}>x_{2}>\cdots>x_{n}>\cdots$.
When $x_{1}x_{1}$ and $x_{2}>x_{3}>\cdots>$ $x_{n}>\cdots$.
Therefore, the constant $a=8^{\frac{1}{8}}$.