Question

if a+b+c=0 then find value of (b+c)^2/bc + ( c+a)^2/ca +(a+b)^2/ab

Answer 1

hope it helps........

Answer 2

Given,

a +b + c = 0

a + b = -c ....................i)

b + c = -a ..............ii)

c + a = -b ..........iii)

Now,

$\frac{(a+b)^{2}}{ab}+\frac{(b+c)^{2}}{bc}+\frac{(c+a)^{2}}{ca}\\\;\\=(\frac{a^{2}+b^{2}+2ab}{ab})+(\frac{b^{2}+c^{2}+2bc}{bc})+(\frac{c^{2}+a^{2}+2ca}{ca})\\\;\\=(\frac{a^{2}}{ab}+\frac{b^{2}}{ab}+\frac{2ab}{ab})+(\frac{b^{2}}{bc}+\frac{c^{2}}{bc}+\frac{2bc}{bc})+(\frac{c^{2}}{ca}+\frac{a^{2}}{ca}+\frac{2ac}{ca})\\\;\\=(\frac{a}{b}+\frac{b}{a}+2)+(\frac{b}{c}+\frac{c}{b}+2)+(\frac{c}{a}+\frac{a}{c}+2)\\\;\\=(\frac{b+c}{a})+(\frac{c+a}{b})+(\frac{a+b}{c})+2+2+2\\\;\\\text{Using eq i),ii) and iii)}\\\;$

$=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}+6\\\;\\=-1-1-1+6\\\;\\=6-3\\\;\\=3$

Hence,

$\frac{(a+b)^{2}}{ab}+\frac{(b+c)^{2}}{bc}+\frac{(c+a)^{2}}{ca}=3$