Question

If $a+b+c=0,$ then the value of $\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{ca}+\frac{(a+b{)}^2}{ab}$

• A.
  0
• B.
  1
• C.
  2
• D.
  3

Answer 1

The correct option is D

3
Given $a+b+c=0\Rightarrow b+c=-a,c+a=-b,a+b=-c\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{ca}+\frac{(a+b{)}^2}{ab}=\frac{(-a{)}^2}{bc}+\frac{(-b{)}^2}{ca}+\frac{(-c{)}^2}{ab}=\frac{a^2}{bc}+\frac{(-b{)}^2}{ca}+\frac{(-c{)}^2}{ab}=\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3$
Hence, the correct answer is option (d)