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
please help!!!(Talking)

Quote:

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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
please help!!!(Talking)

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is that menth to be
or as it is written

Quote:

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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
please help!!!(Talking)

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

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i meant the first exp. not the second one

Quote:

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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

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start by combining the first three terms on the left with the last term using the common denominator