hi everyone,

I have this equation which has sigma notation. I need to expand the sigma into factors and solve for x.

$\displaystyle \underset{abc}{\Sigma}(x-a)[(x-a)^2-(x-b)(x-c)]=0$

my question is did I get the symmetry expansion right? I want to make sure before I try to solve for x. also, in general how do you think about this? I think you have 3 possible combinations in first factor. each of those has 3 combinations of a,b,c. 3+3+3=9 terms in total.

$\displaystyle \\(x-a)[(x-a)^2-(x-b)(x-c)]+\\(x-a)[(x-b)^2-(x-c)(x-a)]+\\(x-a)[(x-c)^2-(x-a)(x-b)]+\\(x-b)[(x-a)^2-(x-b)(x-c)]+\\(x-b)[(x-b)^2-(x-c)(x-a)]+\\(x-b)[(x-c)^2-(x-a)(x-b)]+\\(x-c)[(x-a)^2-(x-b)(x-c)]+\\(x-c)[(x-b)^2-(x-c)(x-a)]+\\(x-c)[(x-c)^2-(x-a)(x-b)]=0$

is this the total expansion?