Algebra Problem

Algebra Level 3

Let $a,b$ and $c$ be non-zero distinct real numbers satisfying $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}.$

Find $|abc|$ .

3 solutions

Laurent Shorts
Apr 4, 2016

$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}$ , $b-c=\frac{c-a}{ca}$ and $c-a=\frac{a-b}{ab}$ . Therefore:

$a-b=\frac{b-c}{bc}=\frac{1}{bc}\frac{c-a}{ca}=\frac{1}{bc·ca}\frac{a-b}{ab}=\frac{a-b}{(abc)^2}$

$(abc)^2=\frac{a-b}{a-b}=1$ ( $a$ and $b$ are distinct) and so $|abc|=1$ .

Shourya Pandey
Apr 12, 2016

Just to add, I think $abc$ is always equal to $-1$ .

Please support it with an explanation..

Ankit Kumar Jain - 3 years, 3 months ago

Chathur Gudesa
Jan 6, 2016

As long as a=b=c , wouldn't this equation be true ? For example , If I substitute a=b=c=2 , wouldn't the answer be 8?

Excuse me if I am mistaken

It is mention in the question that $a$ , $b$ , & $c$ are 3 DISTINCT real numbers.

Ankit Nigam - 5 years, 5 months ago