Prove Equality given a condition

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To prove the equality (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a) under the condition a+b+c=0, the discussion explores the transformation of the terms into u, v, and w, where u+v+w=0. The user simplifies the problem by recognizing that if u+v+w=0, then u^3 + v^3 + w^3 can be expressed as 3uvw. The algebraic expansion of each cubic term leads to a complex expression, which the user struggles to simplify further. Ultimately, the discussion emphasizes the importance of correctly applying polynomial identities to derive the required equality. The conversation concludes with a realization of the relationship between the cubic sums and their product.
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Homework Statement


If a+b+c = 0, show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)[/tex]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> None that I really need to state. <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Well I&#039;ve just messed around with it, grinding through the algebra, and nothing seems to work. I did however notice that the (2a-b), (2b-c) and (2c-a) terms add up to a+b+c, and in this case, 0. So let u=(2a-b), v= (2b-c) and w=(2c-a) so that the question can be simplified to: if u+v+w = 0, then show u^3 + v^3 + w^3 = 3uvw. <br /> <br /> First I need verification if this change is correct, then hints on what I should do next =[
 
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(u+v+w)^3 = 0

Another hint:

Of course (u+v+w)^3 is more profitably expressed as
u^3+v^3+w^3+3u^2(v+w)+3v^2(u+w)+3w^2(u+v)+6uvw
[/color]
 
Yes I got that far with my messing around, but I ran into a wall when trying to see what to do with 3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw :(
 
ah...this looks all roots of polynomial like...I shall help
(A+B)^3=A^3+3A^2B+3AB^2+B^3

now expand out (2a-b)^3 and (2b-c)^3 and (2c-a)^3 and simplify
and then tell me what you get
 
(2a-b)^3 = 8a^3 - 4a^2b + 2ab^2 - b^3

The rest are the same, just replace the letters.

(2a-b)^3 +(2b-c)^3 +(2c-a)^3 = 7a^3 - 4a^2b + 2a^2c + 7b^3 - 4b^2c + 2b^a + 7c^3 - 4c^2a + 2bc^2
 
Gib Z said:
Yes I got that far with my messing around, but I ran into a wall when trying to see what to do with 3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw :(
u+v+w=0 => u+v=-w, and so on.
 
So I get that down to u^3+v^3+w^3 = 2uvw. That is actually quite disturbing...
 
Ahh sorry I forgot about the Original u^3+v^3+w^3 from post 2! Thank you very much~!~!
 
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