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

$$s=\sqrt a + \sqrt b +\sqrt c$$
$$s^2=a+b+c+2(\sqrt {ab} +\sqrt {bc} +\sqrt {ac})$$
I can’t seem to find a way around this obstacle. Squaring it again gives another $\sqrt {ab}+\sqrt {bc} +\sqrt {ac}$, and is seemingly never ending. 
This drove me to the conclusion that the value of s cannot actually be found, and the question must be manipulated to get the required from. I don’t know how to do that though.
 A: How about evaluating the square of $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ independently?
$$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2=ab+bc+ca+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = 11+2s$$
so
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\sqrt{11+2s}$$
From your work, we then get:
$$s^2=9+2\sqrt{11+2s}\Rightarrow s^2-9=2\sqrt{11+2s}\Rightarrow s^4-18s^2+81=44+8s$$
In conclusion $s^4-18s^2-8s=-37$.
A: If we call:
$\rho=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$
we have, squaring s and using Vieta's formulas:
$s^2=9+2\rho$   [1] 
but also, squaring $\rho$:
$\rho^2=11+2s$ [2]
Now take [1] as $s^2-9=2\rho$, square, substitute [2]... and it should be fine :) 
A: You should use the following properties of the roots of a cubic equation ax^3 + bx^2 + cx + d = 0:
a+b+c = -b/a = -(-9)/1 = 9
abc = -d/a = 1
and
ab + ac  + bc = c/a = 11/1 =11
From these you can find:
(a+b+c)^2  =  81 = a^2+b^2+c^2 + 2 (ab + ac + bc) = a^2+b^2+c^2 + 2 * 11
=>  a^2+b^2+c^2 = 81 - 22 = 59
Now expand s^2 and s^4 using these expressions and you should be one step closer.
