Find $s^4-18s^2-8s$ 
Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^4-18s^2-8s$.

$s^4 - 18s^2 - 8s = (s)(s + 4)(s - 2 + \sqrt{6})(s - 2 - \sqrt{6})$
$P(x) = (x - a)(x - b)(x - c)$
From Vieta's formulas:
$a + b+ c = 9$
$abc = 1$
$ab + ac + bc = 11$
Actually, if $y = \sqrt{x}$ then if $y$ is a root, the function must map $y \to x$, which means, let $x \to x^2$ to get:
$P_2(x) = x^6 - 9x^4 + 11x^2 - 1 = 0$. 
Then $s = \sum \text{roots } P_2(x) = 0$.
But then $s^4 - 18s^2 - 8s = 0$? What am I doing wrong?

$$s=\sqrt{a}+\sqrt{b}+\sqrt{c}\\\Rightarrow s^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\\\Rightarrow s^2-9=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\\\Rightarrow s^4-18s^2+81=(s^2-9)^2=4(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2=4(ab+bc+ca)+8\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\\\Rightarrow s^4-18s^2+81=4\cdot11+8\cdot 1\cdot s\\\Rightarrow s^4-18s^2-8s=44-81=-37$$
