2015/105) if $a=xy^2+yz^2+zx^2$ and $b=x^2y+y^2z+z^2x$ then express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in terms of a and b

$(x^3-y^3)(y^3-z^3)(z^3-x^3)$
= $(x^3-y^3)(y^3z^3-y^3x^3-z^6+z^3x^3)$
=$x^3y^3z^3-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6-x^3z^3x^3$
=  $-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6$
= $(z^3x^6+y^3 z^6 + x^3y^6) - (y^3x^6 + x^3z^6 + y^6z^3)$
HENCE
$(x^3-y^3)(y^3-z^3)^(z^3-x^3)$
  $= ((xy^2)^3 + (yz^2)^3+(zx^2)^3) -  ((x^2y)^3 +(y^2z)^3 + (z^2x)^3)\cdots(1)$

we have
$a^3+b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)$
so  $((xy^2)^3 + (yz^2)^3+(zx^2)^3) = (xy^2+yz^2+zx^2)^3- 3(xy^2+yz^2)(yz^2+zx^2)(zx^2+xy^2)$
= $(xy^2+yz^2+zx^2) - 3y(xy+z^2)z(x^2+yz)x(xz+y^2)$
= $b^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(2)$
similarly
$((x^2y)^3 + (y^2z)^3+(z^2x)^3) =a^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(3)$
from (1), (2) and (3) we get

$(x^3-y^3)(y^3-z^3)^(z^3-x^3))=b^3 - a^3$