What is $$\frac{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3}+(z^{2}-x^{2})^{3}}{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}$$

$$(x^2-y^2)+(y^2-z^2)+(z^2-x^2)=0$$

$$\because$$ If $$(a+b+c)=0$$, then $$a^3+b^3+c^3=3abc$$

$$\therefore$$ $$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$$

Similarly, $$(x-y)+(y-z)+(z-x)=0$$

Thus, $$(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-z)(z-x)$$

$$\therefore$$ $$\frac{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3}+(z^{2}-x^{2})^{3}}{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}$$

= $$\frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}$$

= $$\frac{(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)}{(x-y)(y-z)(z-x)}$$

= $$(x+y)(y+z)(z+x)$$

=> Ans - (C)