# 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.

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 $$s^4-18s^2-8s=-37$$.

If we call:

$$\rho=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$

we have, squaring s and using Vieta's formulas:

$$s^2=9+2\rho$$ 

but also, squaring $$\rho$$:

$$\rho^2=11+2s$$ 

Now take  as $$s^2-9=2\rho$$, square, substitute ... and it should be fine :)

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.