6. For any integer $n(n \geqslant 2)$, satisfying $a^{n}=a+1, b^{2 n}=b+3 a$
the size relationship of the positive numbers $a$ and $b$ is ( ).
(A) $a>b>1$
(B) $b>a>1$
(C) $a>1,01$
6.A.

From $a^{n}=a+1>1$, we know $a>1$.
Also, $b^{2 n}=b+3 a>3 a>3$, thus, $b>1$.
Therefore, $a$ and $b$ are both greater than 1.
Since $a^{2 n}-b^{2 n}$
$$
\begin{array}{l}
=(a+1)^{2}-(b+3 a)=a^{2}-a-b+1, \\
\text { and } a^{2 n}-b^{2 n} \\
=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right),
\end{array}
$$

then $a^{2}-a-b+1$
$$
=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right) \text {. }
$$

Therefore, $\frac{a^{2}-a-b+1}{a-b}>1$.
Hence $\frac{a^{2}-a-b+1-a+b}{a-b}>0$, which means
$$
\frac{(a-1)^{2}}{a-b}>0 \text {. }
$$

Thus, $a>b$.
In summary, $a>b>1$.