Implication of equations Just an interesting question I saw online. :)

Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.

Notice that the equality $\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0\leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 \hspace{0.3cm}(1)$ or $a=b=c \hspace{0.3cm}(2)$.


*

*Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
\begin{align}
&\frac{a+b}{c}=-\frac{ab}{c^2}\\
&\frac{c+b}{a}=-\frac{bc}{a^2}\\
&\frac{a+c}{b}=-\frac{ac}{b^2}
\end{align}
$$
 and we finally find that 
$$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=-\frac{ab}{c^2}
-\frac{bc}{a^2}
-\frac{ac}{b^2}=-3$$
by the original equation.


*

*Case(2):
when $a=b=c$ it's easy to see that


$$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
