Identity Crisis

By observation, we know that $1$ is a root of the above equation as,

$(a-b)+(b-c)+(c-a)=0$

But, $1$ does not lie in the interval $(1, 2)$ (as it is an open interval).

Therefore, $2$ more roots lie between $(1, 2)$ .

But, since the given equation is a quadratic, it can only have $1$ more root, not $2$ .

Therefore, since it is a quadratic with more than $2$ roots, it is an identity (This conclusion depends on the fact that there are $2$ roots in the interval, not $\textit{only}$ $2$ ).

If it is an identity, the equation's coefficients must be zero.

$\therefore a-b=b-c=c-a=0$ $\therefore a=b=c$ $\therefore \frac{a^3+b^3+c^3}{abc}=\frac {3a^3}{a^3}=3$