<s>[INST] Find the value of k when $f(x) = x^3 -4x^2 + 6x + k$ If $ \ \ a ,b \ \  $and $\  \ c \ \ $  are roots of $ \ \ f(x) = x^3  -4x^2 + 6x + k \ \   $
and  $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ \ \ $then   find the value of $ \ k $
I have to tell you honestly that I cannot solve this question or find the way to get   value of $ \ k$
I started with   $(a + b +c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \ \ $   $\Rightarrow a^2 + b^2 + c^2 = 4$
then $a^3 +b^3 + c^3 = (a+b+c)(a^2 + b^2 +c^2) - (ab+bc+ac)(a+b+c) + 3abc$
$ \ \ \ \ \ \  \ \ a^3 +b^3 + c^3 =  -8 + 3k$
apply to   $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c)$
$ \ \ \ \ \ \  \ \ \ \ \  \ \ 72 -3k =  3(a+b)(b+c)(a+c)$
finally , I don't know how relation between $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \   $  and $(a+b)(b+c)(a+c)$
Please tell me the way to solve this question correctly.
thank you in advance.
 [/INST] Let $a,b,c$ be the roots of $p(x)=x^3-4x^2+6x+k$. By Viète's theorem we have $a+b+c=4$ and $ab+ac+bc=6$, hence $a^2+b^2+c^2 = (a+b+c)^2-2(ab+ac+bc)=4$ and we know that
$$ \frac{1}{4-a^2}+\frac{1}{4-b^2}+\frac{1}{4-c^2}=1. \tag{1}$$
The following polynomial vanishes over $a^2,b^2,c^2$:
$$ q(x) = -p(\sqrt{x})p(-\sqrt{x}) = x^3-4x^2+(36+8k)x-k^2 \tag{2}$$
and $a^2-4,b^2-4,c^2-4$ are roots of $r(x)=q(x+4)$, i.e. of:
$$ r(x) = x^3+8 x^2+\color{green}{(8k+52)}x-\color{blue}{(k^2-32k-144)}\tag{3}$$
so, always by Viète's theorem,
$$ \frac{1}{4-a^2}+\frac{1}{4-b^2}+\frac{1}{4-c^2} = \frac{-\color{green}{(8k+52)}}{\color{blue}{k^2-32k-144}}=1\tag{4}$$
and that leads to:

$$ k = \color{red}{2\left(6\pm \sqrt{59}\right)}.\tag{4} $$

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