Question 984979
x^3+125=0
factor as sum of cubes
(x+5)(x^2-5x+25)=0
x=-5, one root.
the other has 2 complex roots, conjugates of each other.
x=(1/2)(5+/- sqrt(25-100)) ;; sqrt (-75)=5i sqrt(3)
x=(5/2)+/-(5/2)i sqrt (3)

{{{graph(300,300,-10,10,-500,500,(x^3)+125)}}}