Question

factorise (x-2y)^3+(2y-3z)^3+(3z-x)^3​

Answer 1

Answer:

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see the attachment

thanks

Answer 2

Step-by-step explanation:

We know that,

a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca)

Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc

Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0

Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).