Maybe a simply algeraic question,but i need help~ (1 Viewer)

[Mainoyi]

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

uumm...anyone can help me please?
thanks thanks!!!

 

[YannY]

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

uumm...anyone can help me please?
thanks thanks!!!

haha it dosnt work out lolz it dosnt equate to the answer so quite simply its a wrong question try subbing in a=1 b=-1 c=0

LHS we get 3^3+(-2)^3+1
=20
RHS we get: 3(3)(-2)(-1)
=18

LOLZ????? did you waste hours on a null question?

 

[Slidey]

I tend to agree with YannY: I worked on both the LHS and RHS and they clearly don't match up (the coefficients of b^3 on each side aren't even after elimination of a or c, for example)

Perhaps you should look at this:
http://news.bbc.co.uk/onthisday/hi/dates/stories/june/4/newsid_2496000/2496277.stm
http://www.worldpress.org/Asia/1867.cfm

 

[Mainoyi]

oooooooh...wat the...= =
anywayss...thanks for helping me work out thats a wrong question..

 

[roadrage75]

the questions supposed to be:

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

 

[Slidey]

the questions supposed to be:

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

Are you sure about that? I suspected the third term on LHS might be wrong, but I'm hardly about to try solving it again only to figure out it's wrong.

Hmm. Actually....

Let 2a-b = A, 2b-c = B, 2c-a = C
A+B+C = a+b+c = 0. C = -A-B, sub in
LHS = A^3 + B^3 + C^3 = -3AB(A+B)
RHS = 3ABC = 3AB(-A-B) = -3AB(A+B)

Done.