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.

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.
