If a+b+c= 0, find the value of ((b+c)^2/ bc)+ ((c+a)^2)/ca) + ((a+b)^2/ ab).
Answers
Answer:
Given,
a +b + c = 0
a + b = -c ....................i)
b + c = -a ..............ii)
c + a = -b ..........iii)
Now,
\begin{lgathered}\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)}\\\;\\\end{lgathered}
ab
(a+b)
2
+
bc
(b+c)
2
+
ca
(c+a)
2
=(
ab
a
2
+b
2
+2ab
)+(
bc
b
2
+c
2
+2bc
)+(
ca
c
2
+a
2
+2ca
)
=(
ab
a
2
+
ab
b
2
+
ab
2ab
)+(
bc
b
2
+
bc
c
2
+
bc
2bc
)+(
ca
c
2
+
ca
a
2
+
ca
2ac
)
=(
b
a
+
a
b
+2)+(
c
b
+
b
c
+2)+(
a
c
+
c
a
+2)
=(
a
b+c
)+(
b
c+a
)+(
c
a+b
)+2+2+2
Using eq i),ii) and iii)
\begin{lgathered}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}+6\\\;\\=-1-1-1+6\\\;\\=6-3\\\;\\=3\end{lgathered}
=
a
−a
+
b
−b
+
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
ab
(a+b)
2
+
bc
(b+c)
2
+
ca
(c+a)
2
=3
Answer:
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