+118629 Mmm
ax^2+bx+c, bx^2+cx+a, cx^2+ax+b
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-c \pm \sqrt{c^2-4ab} \over 2b}\\ x = {-a \pm \sqrt{a^2-4cb} \over 2c}\\ \)
These are the zeros for the 3 expressions. Off hand I cannot see why there cannot be 6 roots in total.......
Ok I graphed it in desmos and if does not appear to be able to have 6 roots.
I can get 4 but I could work out how to get more.
https://www.desmos.com/calculator/hrlzvwageo
For any of these to have two roots, the desciminant must be greater than 0.
That can certainly be true for at least one of these but can it be true of 2 or for all there.
Obviously it can be true for 2 cause I graphed it.
So I really do not have a definitive answer. Most likely 4 but I have not seen any maths to prove it.
