<s>[INST] An awful factorisation question. If $a+b+c = 0$ show that $$(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)$$
I have tried substituting the values but it gets too complicated.
Can anyone please help with the method? I have been trying for 30 minutes. 
Thanks!
 [/INST] Let $x=2a-b$, $y=2b-c$, and $z=2c-a$. Then $x+y+z=0$, so $LHS=x^3+y^3+z^3=x^3+y^3+(-x-y)^3=3xy(-x-y)=3xyz=RHS$.
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