prove the identity

• Feb 5th 2010, 06:12 AM
earthboy
prove the identity
if x = (2ab+2bc+2ca)/(a+b+c), then prove that,

[(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3
• Feb 5th 2010, 07:53 AM
Henryt999
Quote:

Originally Posted by earthboy
if x = (2ab+2bc+2ca)/(a+b+c), then prove that,

[(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3

is that menth to be $\displaystyle \frac{6abc}{(x-a)(x-b)(x-c)}$
or as it is written $\displaystyle \frac{6abc}{x-a}\times(x-b)(x-c)$
• Jun 7th 2010, 04:50 PM
earthboy
Quote:

Quote:
Originally Posted by earthboy http://www.mathhelpforum.com/math-he...s/viewpost.gif
if x = (2ab+2bc+2ca)/(a+b+c), then prove that,

[(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3

is that menth to be http://www.mathhelpforum.com/math-he...d44734c1-1.gif
or as it is written http://www.mathhelpforum.com/math-he...eaf3329c-1.gif
i meant the first exp. not the second one
• Jun 7th 2010, 05:03 PM
skeeter
Quote:

Originally Posted by earthboy
if x = (2ab+2bc+2ca)/(a+b+c), then prove that,

[(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3

start by combining the first three terms on the left with the last term using the common denominator $\displaystyle (x-a)(x-b)(x-c)$