Question : What is $\frac{\left (x^{2}-y^{2} \right)^{3}+\left (y^{2}-z^{2} \right )^{3}+\left (z^{2}-x^{2} \right )^{3}}{\left (x-y \right)^{3}+\left (y-z \right )^{3}+\left (z-x \right)^{3}}?$
Option 1: $\frac{(x+y)(y+z)}{(x+z)}$
Option 2: $(x+y)^3(y+z)^3(z+x)^3$
Option 3: $(x+y)(y+z)(z+x)$
Option 4: $(x+y)(y+z)$
Correct Answer: $(x+y)(y+z)(z+x)$
Solution :
$\frac{\left (x^{2}-y^{2} \right )^{3}+\left (y^{2}-z^{2} \right)^{3}+\left (z^{2}-x^{2} \right)^{3}}{\left (x-y \right )^{3}+\left (y-z \right )^{3}+\left (z-x \right)^{3}}$
As we know, if $A + B + C =0$, then $A ^3+ B^3 + C^3=3ABC $
Here, $\left (x^{2}-y^{2} \right )+\left (y^{2}-z^{2} \right)+\left (z^{2}-x^{2} \right)=0$ and $(x-y)+(y-z)+(z-x)=0$
So, $\frac{\left (x^{2}-y^{2} \right )^{3}+\left (y^{2}-z^{2} \right)^{3}+\left (z^{2}-x^{2} \right)^{3}}{\left (x-y \right )^{3}+\left (y-z \right )^{3}+\left (z-x \right)^{3}}$
= $ \frac{3(x^{2}-y^{2}) (y^{2}-z^{2})(z^{2}-x^{2})}{3(x-y)(y-z)(z-x)}$
= $ \frac{(x+y)(y+z)(z+x)(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}$
= $(x+y)(y+z)(z+x)$
Hence, the correct answer is $(x+y)(y+z)(z+x)$.
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