Question 387927
{{{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=-3(x-y)(x+y)(x-z)(x+z)(y-z)(y+z)}}}
{{{(x-y)^3+(y-z)^3+(z-x)^3=-3(x-y)(x-z)(y-z)}}}
So then,
{{{((x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3)/((x-y)^3+(y-z)^3+(z-x)^3)=(-3(x-y)(x+y)(x-z)(x+z)(y-z)(y+z))/(-3(x-y)(x-z)(y-z))}}}
{{{((x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3)/((x-y)^3+(y-z)^3+(z-x)^3)=highlight((x+y)(y+z)(z+x))}}}