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.

We know: $a+b+c=4, ab+bc+ca=6, abc=-k$
Using the condition and $a^2+b^2+c^2=4$ you can get that:
$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}=1=\frac{1}{4-c^2}+\frac{1}{4-a^2}+\frac{1}{4-b^2}\Leftrightarrow -a^2b^2c^2+4(a^2b^2+b^2c^2+c^2a^2)-16(a^2+b^2+c^2)+64=a^2b^2+b^2c^2+c^2a^2-8(a^2+b^2+c^2)+48 \Leftrightarrow a^2b^2c^2-3(a^2b^2+b^2c^2+c^2a^2)+8(a^2+b^2+c^2)-16=0$
Now: $a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=36+8k$
$k^2-3(36+8k)+32-16=0$
$k^2-24k-92=0$
$k_{1,2}=12\pm2\sqrt{59}$
