# 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!

Let $x=2a-b$, $y=2b-c$, and $z=2c-a$. Then $x+y+z=0$, so $LHS=x^3+y^3+z^3=x^3+y^3+(-x-y)^3=3xy(-x-y)=3xyz=RHS$.
Hint: $A^3+B^3=(A+B)(A^2-AB+B^2)$, and $(2a-b)+(2b-c)=2a+b-c = a-2c+(a+b+c)=-(2c-a)$
Use Newton's identities for the polynomial with roots $A=2a-b$, $B=2b-c$, $C=2c-a$. Note that $a+b+c = 0$ implies $A+B+C=0$. So, Newton's identities say that $A^3+B^3+C^3=p_3=e_1 p_2 + e_2 p_1 + 3e_3 = 3e_3 = 3ABC$.
Here, $p_k = A^k + B^k + C^k$, $e_1=p_1=A+B+C$, $e_2=AB+BC+CA$ (but is not needed), and $e_3=ABC$.