Question

If $a+b+c=1$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ where as $a,b,c$ are non zero, then value of $(a+b)ab+(b+c)bc+(c+a)ca$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

$a+b+c=1$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ where $a,b,c$ are $\ne 0$
$\frac{a+b}{c}+1=\frac{1}{c}$
$\frac{b+c}{a}+1=\frac{1}{a}$
$\frac{c+a}{b}+1=\frac{1}{b}$
Adding $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}+3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$
$(a+b)ab+(b+c)bc+(c+a)ca=0$