RMO practice question 7

Since $a,b,c$ are the roots of given equation

By vieta's relations it follows that

$a+b+c=9$

$ab +bc+ca=11$

$abc=1$

Now given $ω= \sqrt{a}+ \sqrt{b}+ \sqrt{c}$

Squaring both sides we obtain

$ω^2=a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$

$\large ω^2=9+2(\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}})$

(Using vieta's relations above)

$\large (ω^2-9)^2=4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}))$

$ω^4-18ω^2+81=4(\frac{ab+bc+ca}{abc}+2(\sqrt{a}+\sqrt{b}+\sqrt{c}))$

$ω^4-18ω^2+81=4(\frac{11}{1}+2ω)$

$ω^4-18ω^2-8ω=44-81=-37$

So $ω^{4}-18ω^{2}-8ω+40=\boxed{3}$

guys. this came to great surprise when my method actually worked. check this https://brilliant.org/discussions/thread/an-easier-way-to-solve-quartics/?ref_id=964870 so basically, $P(x)=x^4-18x^2-8x+37$ we are looking for $p(\omega)+3$ applying my method : $q(x)=x^3+(\frac{-18}{2})x^2+(\frac{18^2-4*37}{16})x -\frac{8^2}{64}=x^3-9x^2+11x-1=0$ that means $p(\omega)+3=0+3=\boxed{3}$