With out finding the cubes,factories the following:
(X-y)³+(y-x)³+(z-x)³.
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Hello user☺☺☺
Since , a+b+c =0
So, a^3 + b^3 + c^3 = 3abc
So, factorization will be...
3(x-y)(y-z)(z-x)
Hope it works
Since , a+b+c =0
So, a^3 + b^3 + c^3 = 3abc
So, factorization will be...
3(x-y)(y-z)(z-x)
Hope it works
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