For any integer $n(n \geqslant 2)$, the positive numbers $a$ and $b$ that satisfy $a^{n}=a+1, b^{2 n}=b+3 a$ have the following relationship:
(A) $a>b>1$
(B) $b>a>1$
(C) $u>1,01$
6. (A).

Given $a>1, b>1$. If $0<1$, then $a>1$, contradiction; if $0<3 a>3$. Thus $b>1$, contradiction. Therefore, $a, b$ are both greater than 1.
On the other hand

Consider $14 i$
Thus
$$
\begin{array}{c}
a^{2 n}-b^{2 n}=(a+1)^{2}-(b+3 a)=a^{2}-a-b+1, \\
a^{2 n}-b^{2 n}=(a-b)\left(a^{2 n-1}+a^{2 n-2} b+\cdots+b^{2 n-1}\right), \\
a^{2}-a-b+1=(a-b)\left(a^{2 n-1}+a^{2 \pi-2} b+\cdots+b^{2 n-1}\right), \\
\frac{a^{2}-a-b+1}{a-b}=a^{2 n}+a^{2 n} b+\cdots+b^{2 n-1}>1,
\end{array}
$$

Therefore, $\frac{a^{2}-2 a+1}{a-b}>0$, i.e., $\frac{(a-1)^{2}}{a-b}>0$. Hence $a>b$.
In conclusion, we have $a>b>1$.