Question

without finding cubes fertilise(x-y)3+(y-z)3+(z-x)3​

Answer 1

When  a+b+c = 0  we know that  a³ + b³ + c³ =  3 a b c

(x-y)³ + (y-z)³ + (z - x)³

     = 3 (x - y) (y -z) (z- x)

     = 3 (xy - x z -y² + yz) (z -x)

     = 3 (xyz - x² y - x z² + x² z - y² z + x y² + y z² - xyz)

     = 3 [ x² (z - y) + z² (y - x)  + y² (x - z) ]

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similarly  x - 2y + 2y - 3z + 3z - x = 0

hence,  (x-2y)³ + (2y - 3z)³ + (3z-x)³ = 3 (x -2y) (2y -3z)(3z - x)

I hope will be right answer