9-7-1. The equation $x^{4}-7 x-3=0$ has exactly two real roots $a$ and $b$, $a>b$. Find the value of the expression $\frac{a-b}{a^{4}-b^{4}}$.
Answer: $1 / 7$.

Solution variant 1. The fact that $a$ and $b$ are roots of the equation simply means that $a^{4}=7 a+3$ and $b^{4}=7 b+3$. Therefore,

$$
\frac{a-b}{a^{4}-b^{4}}=\frac{a-b}{(7 a+3)-(7 b+3)}=\frac{a-b}{7(a-b)}=\frac{1}{7}
$$

As a comment, note that the equation from the condition indeed has two real roots.