Mathematical Olympiad Treasures Problem 1.92 how does this relate to geometric sequence You are supposed to show that if
$a,b,c$ are nonzero real numbers 
and
$(ab+bc+ca)^3 =abc(a+b+c)^3$
$a, b, c$ are terms of a geometric sequence.
One given solution is that:
$(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc)  (b^2 −ac)  (c^2 −ab)$ 
How does the new factorization imply that $a,b,c$ are part of geometric sequence?

You got $(ab+bc+ca)^3 −abc(a+b+c)^3 = (a^2 −bc)  (b^2 −ac)  (c^2 −ab)$. Now since $(ab+bc+ca)^3 =abc(a+b+c)^3$, it implies $(ab+bc+ca)^3 −abc(a+b+c)^3=0$. Then;
$$(a^2 −bc)  (b^2 −ac)  (c^2 −ab)=0$$
This implies one of $(a^2 −bc),(b^2 −ac),(c^2 −ab)$ must be zero. Then we have $a^2 =bc$ or  $b^2 =ac$ or $c^2 =ab$. 
If $a^2=bc$, then $b,a,c$ or $c,a,b$ in G.P. Similar for other cases.
