How to solve this algebraic manipulation problem?

Question

Express the following expression $$E=(x^3-y^3)(y^3-z^3)(z^3-x^3)$$ in terms of $a, b$ where $a,b \in \mathbb R$ and $$a=x^2y+y^2z+z^2x$$ $$b=xy^2+yz^2+zx^2$$

Answer 1

Firstly, we use $x^3-y^3=(x-y)(x^2+xy+y^2)$ for $(x,y)=(a,b),(b,c),(a,c)$ and we obtain $$E=(x-y)(y-z)(z-x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(x^2z+y^2x+z^2y-x^2y-y^2z-z^2x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(b-a)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)=(b-a)E^{'}$$ $$\text{Let's say }c=\frac{a}{xyz}=\frac{x}{z} + \frac {y}{x} + \frac {z}{y} \text{and }d=\frac{b}{xyz}=\frac{x}{y} + \frac {y}{z} + \frac {z}{x}$$ $$\frac{E^{'}}{(xyz)^2}=\prod_{cyc}{(x/yz+1/z+y/xz)}=(\sum_{cyc} \frac{x^2}{y^2} + 2 * \sum_{cyc} \frac{x}{z}) + ( \sum_{cyc} \frac{x^2}{z^2} + 2* \sum_{cyc} \frac{x}{y} ) + (\sum {xy/z^2} +3 ) $$ $$\text{Therefore, }\frac{E^{'}}{(xyz)^2}=c^2 + d^2 + cd=\frac{a^2+b^2+ab}{(xyz)^2}$$ $$\text{Now we have obtained }E^{'}=a^2+b^2+ab\text{, and we know, from earlier, }E=(b-a)E^{'}\text{, therefore}$$ $$E=(b-a)(a^2+b^2+ab) $$ $$E=b^3-a^3 $$

Answer 2

Start: $$ a-b = xy(x-y)+yz (y-z)+zx(z-x) $$ $$= xy(x-y)+yz (y-z)+zx(z-\color{red}y)+zx(\color{red}y-x) $$ $$= (xy-zx)(x-y)+(yz -zx)(y-z)$$

$$ =x(y-z)(x-y)+z(y-x)(y-z) $$ $$ = (x-y)(y-z)(x-z)$$

So $$ E = (a-b)\underbrace{(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)}_{A}$$

Now you have to figer out $A$. (I bet it is $3ab$.)