if a+b+c=0, prove (2a-b)^3 +(2b-c)^3 +(2c-a)^3
= 3 (2a-b)(2b-c)(2c-a)
Do you remember this formula
x^3 +y^3 +z^3 - 3xyz = (x+y+z)(x^2 +y^2 +z^2 - xy-yz-za)
We will use it
when x+y+z=0
Then x^3+y^3+z^3 = 3xyz
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Now in LHS
Consider
2a-b = x
2b -c =y
2c-a= z
Thus x+y+z =
(2a - b) + (2b - c) +(2c -a) = 2(a+b+c) - (a+b+c) = 0
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Hence
x^3 + y^3 +z^3 = (2a-b)^3 +(2b-c)^3 +(2c-a)^3
But using formula
= 3xyz = 3 (2a-b)(2b-c)(2c-a) =RHS
Thus proved