need help with an algebraic question!! (1 Viewer)

[tomcats]

if a+b+c=0, show that (2a-b)^3+(2b-c)^3+(2c-a)^3=3(2a-b)(2b-c)(2c-a)

thx in advance!!!

 

[Archman]

bash!!
lhs = (2a-b+2b-c)((2a-b)^2-(2a-b)(2b-c)+(2b-c)^2)+(2c-a)^3 (sum of cubes)
= (2a+b-c)(...same stuff)+...
= (a-2c)(...)+... (using a+b=-c)
= (a-2c)((2a-b)^2-(2a-b)(2b-c)+(2b-c)^2 - (2c-a)^2 )
= (a-2c)((2a-b+2c-a)(2a-b-2c+a)-(2a-b)(2b-c)+(2b-c)^2) (sum of squares)
= (a-2c)((c-2b)(4a-c)-(2a-b)(2b-c)+(2b-c)^2) (using a+b+c=0 twice)
= (a-2c)((c-2b)(4a-c+2a-b-2b+c))
= (a-2c)(c-2b)(6a-3b)
= 3(2c-a)(2b-c)(2a-b)
= rhs
...yes, crappy question

 

[JamiL]

exactly... not hard but shitty

 

[Abtari]

is this efficient?

i found the question quite tough... but managed to get it finally. dont know if its very efficient tho.

let:
A = 2a-b
B = 2b-c
C = 2c-a

then A+B+C = a+b+c = 0
to prove: A^3 + B^3 + C^3 = 3ABC

since A+B+C = 0
thus cubing both sides, (A+B+C)^3 = 0
expand and voila!

 

[SeDaTeD]

Hehe, and Archman went the bash method.

 

[SaHbEeWaH]

i found the question quite tough... but managed to get it finally. dont know if its very efficient tho.

let:
A = 2a-b
B = 2b-c
C = 2c-a

then A+B+C = a+b+c = 0
to prove: A^3 + B^3 + C^3 = 3ABC

since A+B+C = 0
thus cubing both sides, (A+B+C)^3 = 0
expand and voila!

expanding (A+B+C)³ gives you
A³ + 3BA² + 3AB² + B³ + 3CA² + 6ABC + 3CB² + 3AC² + 3BC² + C³

so A³ + B³ + C³ = -(3BA² + 3AB² + 3CA² + 6ABC + 3CB² + 3AC² + 3BC²)

something looks wrong...

 

[who_loves_maths]

this question is in the inappropriate forum... is it not a year 11 general algebra question ?

i believe, from memory, this is a question from the first chapter of the Year 11 3 unit (red) Cambridge book (the Pender one).

 

[Slidey]

This is the 3unit forum. It is for both preliminary and HSC content.

The primary reason is that HSC students need to know all prelim content anyway, and such content typically isn't 'easier'.

 

[who_loves_maths]

^ my bad then... thought this was a year 12 ext.1 forum, cause there's a year 11 - preliminary forum on its own.

 

[Slidey]

Yeah - and some people ask there questions there. Personally I'd prefer if they asked them here. They will find out more about year 12 maths that way and have a higher chance of getting their question answered.

 

[Abtari]

expanding (A+B+C)³ gives you
A³ + 3BA² + 3AB² + B³ + 3CA² + 6ABC + 3CB² + 3AC² + 3BC² + C³

so A³ + B³ + C³ = -(3BA² + 3AB² + 3CA² + 6ABC + 3CB² + 3AC² + 3BC²)

something looks wrong...

what do you mean?
sorting out the right hand side should give you 3ABC.

 

[SaHbEeWaH]

really?
do show your working..

 

[Slidey]

Looks like Ivan's solution wasn't so bad after all.

 

[FinalFantasy]

Looks like Ivan's solution wasn't so bad after all.

yea man, i reckon someone as high level as Archman would've spotted any simpler solutions if there was.. before doing algebra bash

 

[acmilan]

Heh you got that right FinalFantasy.

If youre ever on here again...Hi Ivan!

 

[Abtari]

really?
do show your working..

let:
A = 2a-b
B = 2b-c
C = 2c-a

then A+B+C = a+b+c = 0
then to prove: A^3 + B^3 + C^3 = 3ABC

since A+B+C = 0
thus cubing both sides, (A+B+C)^3 = 0

A^3 + 3BA^2 + 3AB^2 + B^3 + 3CA^2 + 6ABC + 3CB^2 + 3AC^2 + 3BC^2 + C^3 = 0

so A^3+ B^3 + C^3 = -(3BA^2 + 3AB^2 + 3CA^2 + 6ABC + 3CB^2 + 3AC^2 + 3BC^2)
= -3[AB(A+B) + AC(A+C) + BC(B+C)] - 6ABC
= -3[AB(-C) + AC(-B) + BC(-A)] - 6ABC
= 9ABC - 6ABC
= 3ABC

 

[Slidey]

yea man, i reckon someone as high level as Archman would've spotted any simpler solutions if there was.. before doing algebra bash

It's not that, actually. What I mean is that Ivan's supposed "algebra bash" is short enough and no trickier than Abtari's method. Abtari's is more elegant.