Question

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ where $(a+b+c)\ne 0$ and $abc\ne 0$. What is the value of $(a+b)(b+c)(c+a)?$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$
where $a+b+c\ne 0andabc\ne 0$
$\Rightarrow \frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}$
$\Rightarrow (a+b+c)(bc+ac+ab)=abc=abc$
$\Rightarrow a^{2}c+a^{2}b+b^{2}c+b^{2}a+b{c}^2+a{c}^2+3abc=abc$
$\Rightarrow abc+a^{2}b+a{c}^2+a^{2}c+b^{2}c+b^{2}a+b{c}^2+abc=0$
$\Rightarrow (ab+ac+b^2+bc)(c+a)=0$
$\Rightarrow (a+b)(b+c)(c+a)=0$
Hence option 'A' is correct.