Without finding cubes, factorise (x-y)^{3}+(y-z)^{3}+(z-x)^{3} .

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2. Polynomials

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Without finding the cubes, factorise $(x-y)^{3}+(y-z)^{3}+(z-x)^{3} .$

Option A

Option B

Option C

Option D

Solution

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

Standard 9

Mathematics

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