If $ab+bc+ca=0$
then the value of
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
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(a+b+c)^3 formula ahh – jyothika Aug 27 '14 at 09:31
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What did you try? – user133281 Aug 27 '14 at 09:31
2 Answers
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$$\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+ca+bc}$$ $$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}=\frac{ab+bc+ca}{abc(a+b+c)}=0.$$
mathlove
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I need to know how to edit the question for sin,cos,tan,theta,pi ,summation etc when posting question – jyothika Aug 27 '14 at 09:42
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@DSinghvi: If $a=0$, we have $bc=0$. This leads $a^2-bc=0$, which is a contradiction. So, we have $abc\not =0$. – mathlove Aug 27 '14 at 09:46
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@jyothika: The following page may be helpful : http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – mathlove Aug 27 '14 at 09:48
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@DSinghvi: I suppose $a^2-bc\not =0,b^2-ca\not =0,c^2-ba\not =0$ because the question asks the value of $1/(a^2-bc)+1/(b^2-ca)+1/(c^2-ab)$. – mathlove Aug 27 '14 at 09:50
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$ab+bc+ca=0$
$a(b+c)= -bc$
$b+c= -bc/a$
Adding both side $a$ then $b+c+a=a-bc/a$
$a(a+b+c)= a^2-bc$
$1/(a^2-bc)= 1/a(a+b+c)$
Similarly $1/(b^2-ca) = 1/b(a+b+c)$
and $1/(c^2-ab) = 1/c(a+b+c)$.
By adding all eqution we get
$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$
ASB
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