Without finding the cubes, factorise $(x-y)^{3}+(y-z)^{3}+(z-x)^{3} .$
Without finding the cubes, factorise $(x-y)^{3}+(y-z)^{3}+(z-x)^{3} .$
We know that $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
If $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}-3 x y z=0$ or $x^{3}+y^{3}+z^{3}=3 x y z$
Here, $(x-y)+(y-z)+(z-x)=0$
Therefore, $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)$
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