Alternative arrangement of variables in a polynomial expression

Question

In the course of solving a polynomial equation problem, I seem to have hit a dead-end.

I know that the values of $$(a+b+c)=10 ; abc = 2015 ; ab+bc+ca = 1$$ Now... I've pretty much solved the whole problem (Its mentioning is not of necessity), and all that's left is to find out the value of $$(bc-a^2)(ac-b^2)+(ac-b^2)(ba-c^2)+(ba-c^2)(bc-a^2)$$

I'm not able to figure out the method to factorize the above expression in such a way so as to reduce it to another expression involving $abc, a+b+c$ & $ab+bc+ca$.

How can I proceed?

Answer 1

As in the given,

$a,b,c$ are solutions of the function $x^3-10x^2+x-2015=0$

Hence ${2015\over x}-x^2=1-10x$.

Now

$(bc-a^2)(ac-b^2)+(ac-b^2)(ba-c^2)+(ba-c^2)(bc-a^2)$

$=({2015\over a}-a^2)({2015\over b}-b^2)+({2015\over b}-b^2)({2015\over c}-c^2)+({2015\over c}-c^2)({2015\over a}-a^2)$

$=(1-10a)(1-10b)+(1-10b)(1-10c)+(1-10c)(1-10a)$

$=3-20(a+b+c)+100(ab+bc+ca)$

Answer 2

You can verify that

$$(bc-a^2)(ac-b^2)+(ac-b^2)(ba-c^2)+(ba-c^2)(bc-a^2) = 3(ab+bc+ca)^2-(ab+bc+ca)(a+b+c)^2$$

I'm curious about the original problem.