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