we use the fact that if x+y+z = 0, then x³+y³+ z³ = 3 x y z.
(x²-y²) + (y²-z²) + (z²-x²) = 0
also: (x-y) + (y-z)+ (z-x) = 0
we assume that: x ≠y ≠ z.
hence,
(x²-y²)³ + (y²-z²)³ + (z²-x²)³ ÷ (x-y)³ + (y-z)³ + (z-x)³
= 3 (x²-y²) (y²-z²) (z²-x²) ÷ [3 (x-y) (y-z) (z-x)]
= (x+y) (y+z) (z+x)
