Does the equation
have any solutions for a, b, c, which belong to the set of negative and positive integers?
It helps if you tell us what you have tried or thought about. Otherwise, you learn little from the exercise.
$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} = 1 \implies abc * \left ( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right ) = abc * 1 \implies$
$a^2c + ab^2 + bc^2 = abc.$
$CASE\ I:\ ASSUME\ a, b, c \in \mathbb Z\ and\ a, b, c > 0.$
$a^2c + ab^2 + bc^2 = abc \implies a^2c < abc \implies \dfrac{a^{\cancel {2}}\cancel c}{\cancel a \cancel c} < \dfrac{\cancel a b \cancel c}{\cancel a \cancel c} \implies a < b.$
$Also\ a^2c + ab^2 + bc^2 = abc \implies b^2c < abc \implies \dfrac{b^{\cancel {2}}\cancel c}{\cancel b \cancel c} < \dfrac{a \cancel b \cancel c}{\cancel b \cancel c} \implies b < a.$
$Contradiction \implies assumption\ that\ a, b, c \in \mathbb Z\ and\ a, b, c > 0\ is\ FALSE.$
Now what?
First, if you are going to do this by cases, your second case needs to be a, b, and c all negative.
Second, in your original statement of the problem you did not stipulate that a < b < c. Is that a stipulation or not?
Third if it is NOT a stipulation, you can without loss of generality set up case III as
$a < 0 < b \le c$ and case IV as $a \le b < 0 < c.$
Now all that needs to be done is to figure out which if any of those cases is feasible.
Interesting li'l headacher, Greg.
One thing I'm sure of:
no solution for a,b,c = -999 to 999 skip 0
A quadratic in terms of c gives following restrictions:
(let u = a^2 - ab, v = 4ab^3)
u^2 - v => 0
SQRT(u^2 - v) = integer
-u +- SQRT(u^2 - v) = even
Quadratics in terms of a and b give similar restrictions, of course.