Prove the sequence of three real numbers 
If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then prove that $a,b,c$ are terms in $G.P$

My work:
I assumed that they are in $G.P$ and so assumed $b=ak$ and $c=ak^2$ for some arbitrary $k$. After that I expanded both sides of the equality and got the same results so that means the equality is true. But here's the twist!
Suppose the question was

If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then the terms $a,b,c$ are in $G.P$ or $A.P$ or $H.P\:\:?$

Then what should be one's approach. Will expanding work here$?$
Any help is greatly appreciated.
 A: As mentioned in comments, you haven't really proved that the equation implies $a, b, c$ are in G.P., you have only shown the other way round.
Here is a hint to address the original problem,
$$(ab+bc+ca)^3-abc(a+b+c)^3=(ab-c^2)(bc-a^2)(ca-b^2)$$
Now you can show that if any of the terms on RHS is zero, that implies $a, b, c$ are in G.P.
A: Macavity has given a good answer for how to go about the original problem. To answer your original question,

Suppose the question was
If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then the terms $a,b,c$ are in $G.P$ or $A.P$ or $H.P\:\:?$
Then what should be one's approach. Will expanding work here$?$

You start by trying to come up with counter-examples for each A.P., G.P. H.P, separately. It's not hard to come up with counter-examples for H.P and A.P. But once you struggle to come up with a counter-example for G.P., you might suspect it is true, so maybe you have a go at proving it is true, i.e. that $a,b,c$ are in G.P. . And then if you struggle too much with this, you go back to trying to find a counter-example. Then go back and forth between looking for a counter-example and trying to prove the affirmative. That's the way I do it anyway...
A: We can think of the problem in terms of polynomials too. Assume that $a, b, c$ are the roots of the cubic polynomial '$ P(x) = x^3 + ux^2 + vx + w$. We get
$$-u = a+b+c$$
$$v = ab+bc+ca$$
$$-w = abc$$
Substituting these in the original condition, we have
$$v^3 = (-w)(-u)^3 = wu^3$$ or
$$ w = \frac{v^3}{u^3}$$
$$P(x) = x^3 + ux^2 + vx +\frac{v^3}{u^3} = (x^3 + \frac{v^3}{u^3}) + (ux^2 + vx) $$
$$= (x + \frac{v}{u})(x^2 + \frac{v^2}{u^2} - \frac{v}{u}x) + ux(x+\frac{v}{u}) $$
This shows $\frac{-v}{u}$ is a root of the P(x). The product of three roots is $-w = \frac{-v^3}{u^3}$. So the product of the other two roots is $\frac{v^2}{u^2}$. We get that square of one root is product of other two and they must be in a G.P.
A: Concerning your question, in dealing with relations between three members of a progression for equations such as $ \ (ab+bc+ca)^3 \ = \ abc·(a+b+c)^3 \ \ , \ $ it can be useful to take, say, $ \ b \ $ as the "middle" member and write the other two in terms of it.
For an arithmetic progression with a common difference of $ \ d \  \ , \ $ we have $ \ b - d \ , \ b \ , \ b + d \ \ . \ $ This leads to the putative equation
$$ [ \ (b-d)·b \ + \ b·(b+d) \ + \ (b-d)·(b+d) \ ]^3 \ \ =^{?} \ \ (b-d)·b·(b+d)·(3b)^3 $$
$$ \Rightarrow \ \ [ \ 3b^2  \ - \  d^2 \ ]^3 \ \ \neq \ \ 27·b^4·(b^2 - d^2) \ \ .  $$
We see at once that this is not an equation, since, as a "binomial-cube", the left side must have four terms when "expanded", including powers of $ \ d \ $ as high as $ \ 6 \ \ , \ $ which are clearly not present on the right side.  (The two sides are only equal for $ \ d \ = \ 0 \ \ . \ ] $  The members of a harmonic progression produce a similar situation.  With the members $ \ \frac{1}{\beta - d} \ , \ b = \frac{1}{\beta} \ , \ \frac{1}{\beta + d} \ \ , \ $ we would need to have
$$ [ \ (\beta-d)^{-1}·\beta^{-1} \ + \ \beta^{-1}·(\beta+d)^{-1} \ + \ (\beta-d)^{-1}·(\beta+d)^{-1} \ ]^3 $$ $$ =^{?} \ \ (\beta -d)^{-1}·\beta^{-1}·(\beta+d)^{-1}·[ \ (\beta -d)^{-1}  + \beta^{-1} + (\beta+d)^{-1} \ ]^3 $$
$$ \Rightarrow \ \ \left[ \ \frac{3·\beta}{\beta \ · \ (\beta^2   -   d^2)} \ \right]^3 \ \ \neq \ \ \frac{1}{\beta  ·  (\beta^2   -   d^2)} \  \ · \ \left[ \ \frac{3·\beta^2  \ - \  d^2}{\beta  ·  (\beta^2   -   d^2)} \ \right]^3 \ \ , $$
which is also evidently not an equation for $ \ d \ \neq \ 0 \ \ . $
On the other hand, the members $ \ \frac{b}{r} \ , \ b \ , \ br \ $ of a geometric progression with common ratio $ \ r \ $ produces
$$ \left[ \ \frac{b}{r}·b \ + \ b·br \ + \ \frac{b}{r}·br \ \right]^3 \ \ =^{?} \ \ \left( \ \frac{b}{r} \ · \ b \ · \ br \ \right) \ · \ \left[ \ \frac{b}{r} \ + \ b \ + \ br \ \right]^3 $$
$$ \Rightarrow \ \   (b^2)^3 \ · \ \left[ \ \frac{1}{r}  \ + \ 1 \ + \  r \ \right]^3 \ \ =^{!} \ \ b^3 \ · \ b^3    ·   \left[ \ \frac{1}{r} \ + \ 1 \ + \ r \ \right]^3   $$
as you found by doing something also this line (although we can observe the equality or inequalities without expanding the implied multinomials).
$$ \ \ $$
I agree that Macavity's factorization is probably the most satisfying argument for showing that a geometric progression needs to be present among some ordering of the three numbers in order to satisfy the equation (addressing the original problem).  I had considered dividing out a factor of $ \ abc \ $ from the equation to obtain
$$   \left( \  b \ + \  c \ + \ \frac{b}{a}·c  \ \right)·\left( \ a  \ + \  c \ + \ \frac{a}{b}·c \ \right)·\left( \ a  \ + \  b \ + \ \frac{a}{c}·b \ \right) \ \ = \ \  (a \ + \ b \ + \ c)^3  \ \ . $$
One could make a glib argument that each factor of the left side must match one of the three identical factors on the right side, giving us $ \ \frac{b}{a}·c  \ = \ a \ \Rightarrow \ a^2 \ = \ bc \ \ , \ $ and similarly, $ \ b^2 \ = \ ac \ \ , \  c^2 \ = \ ab \   , \ $ so that a geometric progression must be present among some ordering of $ \ a \ , \ b \ , \ c \ \ . $  However, this is only really convincing if $ \ a + b + c \ $ is a prime integer; otherwise, it is difficult to be persuasive that the equality of the two products  permits us to "match factors" in that way.
