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Factorize : (x-2y)^3+(2y-3z)^3+(3z-x)^3...

Factorize : `(x-2y)^3+(2y-3z)^3+(3z-x)^3`

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Putting `(x-2y)=a,(2y-3z)=b and (3z-x)=c`, we get
`(x-2y)^3+(2y-3z)^3+(3z-x)^3`
`=a^2+b^3+c^3, " where " a+b+c=(x-2y)+(2y-3z)+(3z-x)=0`
`=3abc [because a+b+c=0rArr a^3+b^3+c^3=3abc]`
`=3(x-2y)(2y-3z)(3z-x)`.
`therefore (x-2y)^3+(2y-3z)^3+(3z-x)^3=3(x-2y)(2y-3z)(3z-x)`
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