<s>[INST] 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?
 [/INST] Let us instead see how the cubic whose roots are squares of the roots of $x^3-sx^2+ tx- 1$ (why?) would look like.  Using Vieta, this is easily written as:
$$x^3-(s^2-2t)x^2+(t^2-2s)x-1$$
Comparing with our given cubic, we get $s^2-2t = 9$ and $t^2-2s = 11$ which on eliminating $t$ gives us $s^4 -18s^2-8s = -37$.
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