I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$. I tried for a long time and this is what I got.
$$3abc = a^2 b + b^2 c + c^2 a$$
But it seems like there is a way to determine the value.
On
Substitute $$a+\frac{1}{b}=t,b+\frac{1}{c}=t,c+\frac{1}{a}=t$$ and solve this for $$a,b,c$$ this gives $$a=b=c=\frac{1}{2}(t-\sqrt{t^2-4})$$ or $$a=b=c=\frac{1}{2}(t+\sqrt{t^2-4})$$
On
We start with $$a + \frac 1b = b + \frac 1c = c + \frac 1a$$
This makes three equalities:
\begin{align}a + \frac 1b &= b + \frac 1c\tag{1}\\ b + \frac 1c &= c + \frac 1a\tag{2}\\ c + \frac 1a &= a + \frac 1b\tag{3}\end{align}
First consider equation $(1)$ $$a + \frac 1b = b + \frac 1c$$ which can be rearranged to give \begin{align}a + \frac 1b &= b + \frac 1c\\ a-b&=\frac 1c -\frac 1b\\ &=\frac{b-c}{bc}\end{align}
Similarly, from equation $(2)$, we can see that $$b-c=\frac{c-a}{ac}$$ and thus \begin{align}a-b&=\frac{b-c}{bc}\\ &=\frac{\frac{c-a}{ac}}{bc}\\ &=\frac{c-a}{abc^2}\end{align}
Finally, from equation $(3)$, we get $$c-a=\frac{a-b}{ab}$$ and thus \begin{align}a-b&=\frac{c-a}{abc^2}\\ &=\frac{\frac{a-b}{ab}}{abc^2}\\ &=\frac{a-b}{a^2b^2c^2}\end{align}
We can now rearrange this as follows (when $a\neq b$) \begin{align}a-b&=\frac{a-b}{a^2b^2c^2}\\ (a-b)(a^2b^2c^2)&=a-b\\ a^2b^2c^2&=\frac{a-b}{a-b}\\ a^2b^2c^2&=1\\ \sqrt{a^2b^2c^2}&=\sqrt1\\ \sqrt{a^2}\sqrt{b^2}\sqrt{c^2}&=\pm1\tag{$*$}\\ abc&=\pm1\end{align}
where $(*)$ comes from the fact that $\sqrt{a\cdot b} =\sqrt{a}\cdot\sqrt{b}$
On
If $a=b$ then $b=c$ and $a=b=c$, which says that $abc$ is not defined.
Let $(a-b)(a-c)(b-c)\neq0.$
Thus, since $$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc},$$ $$b-c=\frac{c-a}{ac}$$ and $$c-a=\frac{a-b}{ab},$$ we obtain $$(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2},$$ which gives $$abc=1$$ or $$abc=-1.$$
Hint
\begin{align} a-b&= \frac1c-\frac 1b \\ &=\frac{b-c}{bc}\\ &=\frac{c-a}{abc^2} \\& =\frac{a-b}{a^2b^2c^2}. \end{align}