Asymmetric Polynomial II

Let $X = a^2b +b^2c + c^2a$ and $Y = ab^2 + bc^2 + ca^2$ .

$\Rightarrow X - Y = ab(a-b) + bc(b-c) + ca(c-a)$

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

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

$= (a-c)(a-b)(b-c)$

Now, using the definition of discriminant we have

$\Delta = (a-c)^2(a-b)^2(b-c)^2$

The discriminant for the cubic polynomial $px^3 + qx^2 + rx + s$ is given by

$q^2r^2 -4pr^3 -4q^3s - 27p^2s^2 + 18pqrs$

On Solving we get $\Delta = 765625$

Since $X - Y$ is positive we have $X - Y$ = $\sqrt{\Delta} = 875$

Also $(a + b+ c)(ab +bc+ac) = X + Y + 3abc$

Using Vieta's Formula we get

$X + Y = 125$

Solving the 2 equations we get $X = 500$ .

$a+b+c=10, ab+bc+ca=-25, abc = -125$ .

Terms like $a^{2}b, b^{2}c, c^{2}a$ can be got by multiplying $(a+b+c)$ and $(ab+bc+ca)$ .
$(a+b+c)(ab+bc+ca)=a^{2}b+b^{2}c+c^{2}a + ab^{2}+bc^{2}+ca^{2}+3abc$ .

Let $X = a^{2}b+b^{2}c+c^{2}a$ and $Y = ab^{2}+bc^{2}+ca^{2}$ .
Straightforward algebraic manipulations give
$X+Y = (a+b+c)(ab+bc+ca) - 3abc$ . Inserting the values, $X+Y=125$ .

Also, $X-Y=(a-b)(a-c)(b-c)$ . Using the given constraint $a > b> c$ ,
$X-Y > 0$ i.e. $X> Y$ .

Multiplying $X$ & $Y$ ,
$XY=a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}+abc(a^{3}+b^{3}+c^{3})+3(abc)^{2}$ .

It can be shown that
$a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3} = (ab+bc+ca)^{3}-3abc(a+b+c)(ab+bc+ca)+3(abc)^{2}$ .
$a^{3}+b^{3}+c^{3} = (a+b+c)^{3}-3(a+b+c)(ab+bc+ca)+3abc$ .
Plugging the values, $XY=-12\times125^{2}$ .

$X+Y=125$ and $XY=-12\times125^{2}$ .
Solving $X=4\times125, Y=-3\times125$ .
$X=500$ .