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.

Let $\alpha^3-U\alpha^2+V\alpha-W$ have roots $x$, $y$, $z$, then
$U=x+y+z$ $V=xy+xz+yz$ and $W=xyz$. Notice that:
$$
\frac{1}{x^2+y^2}+\frac{1}{x^2+z^2}+\frac{1}{y^2+z^2}=\frac{x^4+3 x^2 y^2+3 x^2 z^2+y^4+3 y^2 z^2+z^4}{\left(x^2+y^2\right) \left(x^2+z^2\right) \left(y^2+z^2\right)}
$$
And:
$$
x^4 + 3 x^2 y^2 + y^4 + 3 x^2 z^2 + 3 y^2 z^2 + z^4=(x+y+z)^4-4 (x y+x z+y z) (x+y+z)^2-2 x y z (x+y+z)+5 (x y+x z+y z)^2=
$$
$$
=U^4-4VU^2+2WU+5V^2
$$
$$
(x^2 + y^2) (x^2 + z^2) (y^2 + z^2)=
$$
$$
-x^2 y^2 z^2-2 x y z (x+y+z)^3+(x y+x z+y z)^2 (x+y+z)^2+4 x y z (x y+x z+y z) (x+y+z)-2 (x y+x z+y z)^3=
$$
$$
=-2 U^3 W+U^2 V^2+4 U V W-2 V^3-W^2
$$
These identities come from the general technique. (Look up elementary symmetric polynomials)
So we have:
$$
\frac{U^4-4 U^2 V-2 U W+5 V^2}{-2 U^3 W+U^2 V^2+4 U V W-2 V^3-W^2}=\beta
$$
Which should be solved for $W$. The soluton is:
$$
W=\frac{2 \beta  U V+U-\beta  U^3\pm\sqrt{\beta^2 \left(U^2-2 V\right) \left(U^2-V\right)^2+U^2+\beta  \left(-3 U^4+8 U^2 V-5 V^2\right)}}{\beta }
$$
In your case, this gives $-12\pm2\sqrt{59}$.
