Question

If $abc=1$, then the value of $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$ is

• A. $abc$
• B. $a$
• C. $1$
• D. $3$

Answer 1

Answer: (A), (C)

$abc$
$1$

To find $\to$ $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$

$abc=1\to a=\frac{1}{bc}$

So $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=$

$=\frac{1}{1+\frac{1}{bc}+b^{-1}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+bc}$

$\Rightarrow \frac{bc}{1+c+bc}+\frac{c}{1+c+bc}+\frac{1}{bc+c+1}=1$