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**M.R** Hi, I am attempting the following question, but am stuck:

If $\displaystyle a+b+c=0$ , show that $\displaystyle (2a-b)^3+(2b-c)^3+(2c-a)^3 = 3(2a-b)(2b-c)(2c-a)$

Here is what i have tired:

$\displaystyle (2a-b)^3=8a^3-12a^2b+6ab^2-b^3$

$\displaystyle (2b-c)^3=8b^3-12b^2c+6bc^2-c^3$

$\displaystyle (2c-a)^3=8c^3-12c^2a+6ca^2-a^3$

Adding all those together i get:

$\displaystyle =8(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)-(a^3+b^3+c^3)$

$\displaystyle =7(a^3+b^3+c^3)-12(a^2b+b^2c+c^2a)+6(ab^2+bc^2+ca^2)$

And then i don't know what to do...