Question

If $a+b+c=0$, then $\frac{(a+b{)}^2}{ab}+\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{ca}$ is equal to

• A. $3$
• B. $3abc$
• C. $3(a+b+c)$
• D. $3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

Answer 1

The correct option is A $3$

a+b+c = 0
$\frac{(a+b{)}^2}{ab}+\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{bc}$
a+b =-c
b+c=-a
c+a=-b
then $\frac{(c{)}^2}{ab}+\frac{(a{)}^2}{bc}+\frac{(b{)}^2}{ca}$
$\frac{a^2+b^3+c^3}{abc}=\frac{3\left(abc\right)}{abc}=3$