$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$ Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$

I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it gave a factorised form.
$$(ab+bc+ca).((ab)^2+(bc)^2+(ca)^2-abc.(a+b+c))=abc.(a+b+c).(a^2+b^2+c^2-ab-bc-ca)$$
How can I proceed after that?


There was a good answer posted by @Albusdumbledore that was deleted. I am just  reproducing that answer and will delete mine if the other is undeleted.
The suggestion was to write $a = q, b = q r$ and $c = q r k$. Naturally, since $a,b,c \ne 0$, we also have $q,r,k \ne 0$. Plugging this into the relation, one concludes that $r, k$ must satisfy the relation
$$
(k-r) \left(k^2 r-1\right) \left(k r^2-1\right)=0
$$
All solutions lead to GP's. More specifically,
Case 1. If $k=r$, $a=q$, $b = r q$ and $c= r^2 q$ are consecutive terms in a GP with ratio $r$.
Case 2. If $k=\frac{1}{r^2}$ then $c=\frac{1}{r^2} q$, $a = q$ and $b=rq$ are (nonconsecutive) terms in a GP with ratio $r$
Case 3. If $k=\frac{1}{\sqrt{r}}$ then $c= \frac 1r q$, $a=q$ and $b= rq$ are (nonconsecutive) terms in a GP with ratio $\sqrt{r}$.
Case 4. If $k=-\frac{1}{\sqrt{r}}$ then $a=q$, $c=-\sqrt{r} q$, $b = rq$ are consecutive terms in a GP with ratio $-\sqrt{r}$
