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Factorize : (x-y)^3+(y-z)^3+(z-x)^3...

Factorize : `(x-y)^3+(y-z)^3+(z-x)^3`

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We know the corollary: if `a+b+c=0` then `a^3+b^3+c^3=3abc`
Using the above identity taking `a=x−y, b=y−z and c=z−x,` we have `a+b+c=x−y+y−z+z−x=0` then the equation `(x−y)^3+(y−z)^3+(z−x)^3`can be factorised as follows:
`(x−y)^3+(y−z)^3+(z−x)^3=3(x−y)(y−z)(z−x)`
Hence, `(x−y)^3+(y−z)^3+(z−x)^3=3(x−y)(y−z)(z−x)`
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