<s>[INST] $(ab+bc+ca)^3=abc(a+b+c)^3$, 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?

 [/INST] This one is a long-way of solving (With a very obvious result).
\begin{gather*}
\implies(ab+bc+ca)^3-abc(a+b+c)^3=0\\
\end{gather*}
\begin{multline*}
\implies (a^3b^3+b^3c^3+c^3a^3+3a^2b^3c+3a^2bc^3+3a^3b^2c+3ab^2c^3+3a^3bc^2+3ab^3c^2)-\\abc(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2)=0
\end{multline*}
After simplifying
\begin{equation}
\implies a^3b^3+b^3c^3+c^3a^3-a^4bc-ab^4c-abc^4=0
\end{equation}
Factorising this will give
\begin{equation}
\implies(ab-c^2)(bc-a^2)(ca-b^2)=0
\end{equation}
Therefore, a,b,c must be in G.P.
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