Cyclic polynomial proof 2. So here's my second question for the day, 
Q. Manipulate the given equality
$$a^2b^2(bc-a^2)+b^2c^2(ca-b^2)+c^2a^2(ab-c^2)$$$$ = a^2b^2(b^2-ac)+b^2c^2(c^2-ab)+c^2a^2(a^2-bc)$$
to obtain the following:
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)=abc(a+b)(b+c)(c+a)$$

Rearrange the LHS as
$$a^2b^2(bc)+b^2c^2(ca)+c^2a^2(ab) - (a^2b^2(a^2)+b^2c^2(b^2)+c^2a^2(c^2)) = A - B$$
Rearrange the RHS as
$$a^2b^2(b^2)+b^2c^2(c^2)+c^2a^2(a^2) - (a^2b^2(ac)+b^2c^2(ab)+c^2a^2(bc)) = C - D$$
So now we have
$$A - B = C - D$$
Exchange
$$A + D  = B + C$$
First we have the following
$$(x+y)(y+z)(x+z) = x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz$$
We have
$$A+D = a^2b^2(bc)+b^2c^2(ca)+c^2a^2(ab) + a^2b^2(ac)+b^2c^2(ab)+c^2a^2(bc)$$
$$= (a^2 b^2+c^2a^2)(bc)+(a^2b^2+b^2c^2)(ca)+(c^2a^2+b^2c^2)(ab)$$
$$= abca(b^2+c^2) + abcb(a^2+c^2) + abcc ( a^2+b^2)$$
$$= abc(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)$$
$$= abc(ab^2+a^2b+a^2c+ac^2+b^2c+bc^2)$$
$$= abc((a+b)(b+c)(a+c)-2abc)$$
$$= abc (a+b)(b+c)(a+c) - 2(abc)^2$$
On the other hand
$$B + C = a^2b^2(a^2)+b^2c^2(b^2)+c^2a^2(c^2) + 
a^2b^2(b^2)+b^2c^2(c^2)+c^2a^2(a^2)$$
$$= a^2b^2(a^2+b^2)+b^2c^2(b^2+c^2)+c^2a^2(a^2+c^2)$$
$$= a^4b^2 + a^2 b^4 + b^4c^2 + b^2c^4 + a^4 c^2 + a^2c^4$$
$$= (a^2+b^2)(b^2+c^2)(a^2+c^2) - 2 a^2 b^2 c^2$$
By cancelling the $2a^2b^2c^2$ term from each side you have the required identity.
