Prove Equality given a condition

  • Thread starter Thread starter Gib Z
  • Start date Start date
  • Tags Tags
    Condition
Click For Summary

Homework Help Overview

The problem involves proving an equality under the condition that a + b + c = 0. The specific expression to be shown is (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a).

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss manipulating the expression by substituting u, v, and w for the terms (2a-b), (2b-c), and (2c-a), respectively. There is an exploration of the identity u + v + w = 0 and its implications for the cubic terms. Some participants express uncertainty about how to proceed after expanding the cubic expressions.

Discussion Status

The discussion is ongoing, with participants sharing their algebraic manipulations and insights. There is a recognition of the relationship between the terms and the condition a + b + c = 0, but no consensus has been reached on the next steps or the final outcome.

Contextual Notes

Some participants note the challenge of simplifying the expanded forms and the complexity introduced by the cubic terms. There is a mention of potential confusion regarding the algebraic identities involved.

Gib Z
Homework Helper
Messages
3,341
Reaction score
7

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]<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'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 [itex]u^3 + v^3 + w^3 = 3uvw[/itex]. <br /> <br /> First I need verification if this change is correct, then hints on what I should do next =[[/itex]
 
Physics news on Phys.org
(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 [itex]3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw[/itex] :(
 
ah...this looks all roots of polynomial like...I shall help
[tex](A+B)^3=A^3+3A^2B+3AB^2+B^3[/tex]

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

The rest are the same, just replace the letters.

[tex](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[/tex]
 
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 [itex]3u^2 (v+w) +3v^2 (u+w) +3w^2 (u+v) + 6uvw[/itex] :(
u+v+w=0 => u+v=-w, and so on.
 
So I get that down to [tex]u^3+v^3+w^3 = 2uvw[/tex]. 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~!~!
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K