6. Let $a, b, c$ be the roots of $x^{3}-9 x^{2}+11 x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^{4}-18 s^{2}-8 s$.
Answer: -37
Solution: First of all, as the left side of the first given equation takes values $-1,2$, -7 , and 32 when $x=0,1,2$, and 3 , respectively, we know that $a, b$, and $c$ are distinct positive reals. Let $t=\sqrt{a b}+\sqrt{b c}+\sqrt{c a}$, and note that
$$
\begin{aligned}
s^{2} & =a+b+c+2 t=9+2 t, \\
t^{2} & =a b+b c+c a+2 \sqrt{a b c} s=11+2 s, \\
s^{4} & =(9+2 t)^{2}=81+36 t+4 t^{2}=81+36 t+44+8 s=125+36 t+8 s, \\
18 s^{2} & =162+36 t,
\end{aligned}
$$
so that $s^{4}-18 s^{2}-8 s=-37$.