Prove Equality given a condition

[Gib Z]

Homework Statement

If a+b+c = 0, show that [itex](2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)[/tex]

Homework Equations

None that I really need to state.

The Attempt at a Solution

Well I'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 $u^3 + v^3 + w^3 = 3uvw$.

First I need verification if this change is correct, then hints on what I should do next =[

 

[morphism]

(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

 

[Gib Z]

Yes I got that far with my messing around, but I ran into a wall when trying to see what to do with $3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw$ :(

 

[rock.freak667]

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

 

[Gib Z]

$$(2a-b)^3 = 8a^3 - 4a^2b + 2ab^2 - b^3$$

The rest are the same, just replace the letters.

$$(2a-b)^3 +(2b-c)^3 +(2c-a)^3 = 7a^3 - 4a^2b + 2a^2c + 7b^3 - 4b^2c + 2b^a + 7c^3 - 4c^2a + 2bc^2$$

 

[morphism]

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.

 

[Gib Z]

So I get that down to $$u^3+v^3+w^3 = 2uvw$$. That is actually quite disturbing...

 

[Gib Z]

Ahh sorry I forgot about the Original u^3+v^3+w^3 from post 2! Thank you very much~!~!