<s>[INST] 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.
 [/INST] 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.
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